# Lessons from teaching gifted elementary school students: Index (updated)

I’m doing something that I should have done a long time ago: collect past series of posts into a single, easy-to-reference post. The following posts formed my series on various lessons I’ve learned while trying to answer the questions posed by gifted elementary school students. (This is updated from my previous index.)

Part 1: A surprising pattern in some consecutive perfect squares.

Part 2: Calculating 2 to a very large exponent.

Part 3a: Calculating 2 to an even larger exponent.

Part 3b: An analysis of just how large this number actually is.

Part 4a: The chance of winning at BINGO in only four turns.

Part 4b: Pedagogical thoughts on one step of the calculation.

Part 4c: A complicated follow-up question.

Part 5a: Exponentiation is multiplication as multiplication is to addition. So, multiplication is to addition as addition is to what? (I offered the answer of incrementation, but it was rejected: addition requires two inputs, while incrementation only requires one.)

Part 5b: Why there is no binary operation that completes the above analogy.

Part 5c: Knuth’s up-arrow notation for writing very big numbers.

Part 5d: Graham’s number, reputed to be the largest number ever to appear in a mathematical proof.

Part 6a: Calculating $(255/256)^x$.

Part 6b: Solving $(255/256)^x = 1/2$ without a calculator.

Part 7a: Estimating the size of a 1000-pound hailstone.

Part 7b: Estimating the size a 1000-pound hailstone.

Part 8a: Statement of an usually triangle summing problem.

Part 8b: Solution using binomial coefficients.

Part 8c: Rearranging the series.

Part 8d: Reindexing to further rearrange the series.

Part 8e: Rewriting using binomial coefficients again.

Part 8f: Finally obtaining the numerical answer.

Part 8g: Extracting the square root of the answer by hand.

# Lessons from teaching gifted elementary school students: Index (updated)

I’m doing something that I should have done a long time ago: collect past series of posts into a single, easy-to-reference post. The following posts formed my series on various lessons I’ve learned while trying to answer the questions posed by gifted elementary school students. (This is updated from my previous index.)

Part 1: A surprising pattern in some consecutive perfect squares.

Part 2: Calculating 2 to a very large exponent.

Part 3a: Calculating 2 to an even larger exponent.

Part 3b: An analysis of just how large this number actually is.

Part 4a: The chance of winning at BINGO in only four turns.

Part 4b: Pedagogical thoughts on one step of the calculation.

Part 4c: A complicated follow-up question.

Part 5a: Exponentiation is multiplication as multiplication is to addition. So, multiplication is to addition as addition is to what? (I offered the answer of incrementation, but it was rejected: addition requires two inputs, while incrementation only requires one.)

Part 5b: Why there is no binary operation that completes the above analogy.

Part 5c: Knuth’s up-arrow notation for writing very big numbers.

Part 5d: Graham’s number, reputed to be the largest number ever to appear in a mathematical proof.

# Lessons from teaching gifted elementary school students: Index

I’m doing something that I should have done a long time ago: collect past series of posts into a single, easy-to-reference post. The following posts formed my series on various lessons I’ve learned while trying to answer the questions posed by gifted elementary school students.

Part 1: A surprising pattern in some consecutive perfect squares.

Part 2: Calculating 2 to a very large exponent.

Part 3a: Calculating 2 to an even larger exponent.

Part 3b: An analysis of just how large this number actually is.

Part 4a: The chance of winning at BINGO in only four turns.

Part 4b: Pedagogical thoughts on one step of the calculation.

Part 4c: A complicated follow-up question.

# Lessons from teaching gifted elementary school students (Part 3b)

Every so often, I’ll informally teach a class of gifted elementary-school students. I greatly enjoy interacting with them, and I especially enjoy the questions they pose. Often these children pose questions that no one else will think about, and answering these questions requires a surprisingly depth of mathematical knowledge.

Here’s a question I once received:

Suppose

$A \times A = B$

$B \times B \times B = C$

$C \times C \times C \times C= D$

If the pattern goes on, and if $A = 2$, what is $Z$?

In yesterday’s post, we found that the answer was

$Z =2^{26!} = 10^{26! \log_{10} 2} \approx 10^{1.214 \times 10^{26}}$,

a number with approximately $1.214 \times 10^{26}$ digits.

How can we express this number in scientific notation? We need to actually compute the integer and decimal parts of $26! \log_{10} 2$, and most calculators are not capable of making this computation.

Fortunately, Mathematica is able to do this. We find that

$Z \approx 10^{121,402,826,794,262,735,225,162,069.4418253767}$

$\approx 10^{0.4418253767} \times 10^{121,402,826,794,262,735,225,162,069}$

$\approx 2.765829324 \times 10^{121,402,826,794,262,735,225,162,069}$

Here’s the Mathematica syntax to justify this calculation. In Mathematica, $\hbox{Log}$ means natural logarithm:

Again, just how big is this number? As discussed yesterday, it would take about 12.14 quadrillion sheets of paper to print out all of the digits of this number, assuming that $Z$ was printed in a microscopic font that uses 100,000 characters per line and 100,000 lines per page. Since 250 sheets of paper is about an inch thick, the volume of the 12.14 quadrillion sheets of paper would be

$1.214 \times 10^{16} \times 8.5 \times 11 \times \displaystyle \frac{1}{250} \hbox{in}^3 \approx 1.129 \times 10^{17} \hbox{in}^3$

By comparison, assuming that the Earth is a sphere with radius 4000 miles, the surface area of the world is

$4 \pi (4000 \times 5280 \times 12) \hbox{in}^2 \approx 8.072 \times 10^{17} \hbox{in}^2$.

Dividing, all of this paper would cover the entire world with a layer of paper about $0.14$ inches thick, or about 35 sheets deep. In other words, the whole planet would look something like the top of my desk.

What if we didn’t want to print out the answer but just store the answer in a computer’s memory? When written in binary, the number $2^{26!}$ requires…

$26!$ bits of memory, or…

about $4.03 \times 10^{26}$ bits of memory, or…

about $latex 5.04 \times 10^{25} bytes of memory, or … about $5.04 \times 10^{13}$ terabytes of memory, or… about 50.4 trillion terabytes of memory. Suppose that this information is stored on 3-terabyte external hard drives, so that about $50.4/3 = 16.8$ trillion of them are required. The factory specs say that each hard drive measures $129 \hbox{mm} \times 42 \hbox{mm} \times 167 \hbox{mm}$. So the total volume of the hard drives would be $1.52 \times 10^{19} \hbox{mm}^3$, or $15.2 \hbox{km}^3$. By way of comparison, the most voluminous building in the world, the Boeing Everett Factory (used for making airplanes), has a volume of only $0.0133 \hbox{km}^3$. So it would take about 1136 of these buildings to hold all of the necessary hard drives. The cost of all of these hard drives, at$100 each, would be about $1.680 quadrillion. So it’d be considerably cheaper to print this out on paper, which would be about one-seventh the price at$242 trillion.

Of course, a lot of this storage space would be quite repetitive since $2^{26!}$, in binary, would be a one followed by $26!$ zeroes.

# Lessons from teaching gifted elementary school students (Part 3a)

Every so often, I’ll informally teach a class of gifted elementary-school students. I greatly enjoy interacting with them, and I especially enjoy the questions they pose. Often these children pose questions that no one else will think about, and answering these questions requires a surprisingly depth of mathematical knowledge.

Here’s a question I once received:

Suppose

$A \times A = B$

$B \times B \times B = C$

$C \times C \times C \times C= D$

If the pattern goes on, and if $A = 2$, what is $Z$?

I leave a thought bubble in case you’d like to think this. (This is significantly more complicated to do mentally than the question posed in yesterday’s post.) One way of answering this question appears after the bubble.

Let’s calculate the first few terms to try to find a pattern:

$B = 2 \times 2 = 2^2$

$C = 2^2 \times 2^2 \times 2^2 = 2^6$

$D = 2^6 \times 2^6 \times 2^6 \times 2^6 = 2^{24}$

etc.

Written another way,

$A = 2^1 = 2^{1!}$

$B = 2^{2!}$

$C = 2^{3!}$

$D = 2^{4!}$

Naturally, elementary school students have no prior knowledge of the factorial function. That said, there’s absolutely no reason why a gifted elementary school student can’t know about the factorial function, as it only consists of repeated multiplication.

Continuing the pattern, we see that $Z = 2^{26!}$. Using a calculator, we find $Z \approx 2^{4.032014611 \times 10^{26}}$.

If you try plugging that number into your calculator, you’ll probably get an error. Fortunately, we can use logarithms to approximate the answer. Since $2 = 10^{\log_{10} 2}$, we have

$Z = \left( 10^{\log_{10} 2} \right)^{4.032014611 \times 10^{26}} = 10^{4.032014611 \times 10^{26} \log_{10} 2}$

Plugging into a calculator, we find that

$Z \approx 10^{1.214028268 \times 10^{26}} = 10^{121.4028628 \times 10^{24}}$

We conclude that the answer has more than 121 septillion digits.

How big is this number? if $Z$ were printed using a microscopic font that placed 100,000 digits on a single line and 100,000 lines on a page, it would take 12.14 quadrillion pieces of paper to write down the answer (6.07 quadrillion if printed double-sided). If a case with 2500 sheets of paper costs $100, the cost of the paper would be$484 trillion ($242 trillion if double-sided), dwarfing the size of the US national debt (at least for now). Indeed, the United States government takes in about$3 trillion in revenue per year. At that rate, it would take the country about 160 years to raise enough money to pay for the paper (80 years if double-sided).

And that doesn’t even count the cost of the ink or the printers that would be worn out by printing the answer!

# Lessons from teaching gifted elementary school students (Part 2)

Every so often, I’ll informally teach a class of gifted elementary-school students. I greatly enjoy interacting with them, and I especially enjoy the questions they pose. Often these children pose questions that no one else will think about, and answering these questions requires a surprisingly depth of mathematical knowledge.

Here’s a question I once received:

Suppose

$A \times A = B$

$B \times B = C$

$C \times C = D$

If the pattern goes on, and if $A = 2$, what is $Z$?

I leave a thought bubble in case you’d like to think this. One way of answering this question appears after the bubble.

Let’s calculate the first few terms to try to find a pattern:

$B = 2 \times 2 = 2^2$

$C = 2^2 \times 2^2 = 2^4$

$D = 2^4 \times 2^4 = 2^8$

etc.

Written another way,

$A = 2^1 = 2^{2^0}$

$B = 2^{2^1}$

$C = 2^{2^2}$

$D = 2^{2^3}$

Continuing the pattern, we see that $Z = 2^{2^{25}}$, or $Z = 2^{33,554,432}$.

If you try plugging that number into your calculator, you’ll probably get an error. Fortuniately, we can use logarithms to approximate the answer. Since $2 = 10^{\log_{10} 2}$, we have

$Z = \left( 10^{\log_{10} 2} \right)^{33,554,432} = 10^{33,554,432 \log_{10} 2}$

Plugging into a calculator, we find that

$Z \approx 10^{10,100,890.5195}$

$\approx 10^{0.5195} 10^{10,100,890}$

$\approx 3.307 \times 10^{10,100,890}$

When this actually happened to me, it took me about 10 seconds to answer — without a calculator — “I’m not sure, but I do know that the answer has about 10 million digits.” Naturally, my class was amazed. How did I do this so quickly? I saw that the answer was going to be $Z = 2^{2^{25}}$, so I used the approximation $2^{10} \approx 1000$ to estimate

$2^{25} = 2^5 \times 2^{10} \times 2^{10} \approx 32 \times 1000 \times 1000 = 32,000,000$

Next, I had memorized the fact that that $\log_{10} 2 \approx 0.301 \approx 1/3$. So I multiplied $32,000,000$ by $1/3$ to get approximately 10 million. As it turned out, this approximation was a lot more accurate than I had any right to expect.

# Square roots and logarithms without a calculator (Part 6)

I’m in the middle of a series of posts concerning the elementary operation of computing a square root. This is such an elementary operation because nearly every calculator has a $\sqrt{~~}$ button, and so students today are accustomed to quickly getting an answer without giving much thought to (1) what the answer means or (2) what magic the calculator uses to find square roots. I like to show my future secondary teachers a brief history on this topic… partially to deepen their knowledge about what they likely think is a simple concept, but also to give them a little appreciation for their elders.

In Parts 3-5 of this series, I discussed how log tables were used in previous generations to compute logarithms and antilogarithms.

Today’s topic — log tables — not only applies to square roots but also multiplication, division, and raising numbers to any exponent (not just to the $1/2$ power). After showing how log tables were used in the past, I’ll conclude with some thoughts about its effectiveness for teaching students logarithms for the first time.

To begin, let’s again go back to a time before the advent of pocket calculators… say, the 1880s.

Aside from a love of the movies of both Jimmy Stewart and John Wayne, I chose the 1880s on purpose. By the end of that decade, James Buchanan Eads had built a bridge over the Mississippi River and had designed a jetty system that allowed year-round navigation on the Mississippi River. Construction had begun on the Panama Canal. In New York, the Brooklyn Bridge (then the longest suspension bridge in the world) was open for business. And the newly dedicated Statue of Liberty was welcoming American immigrants to Ellis Island.

And these feats of engineering were accomplished without the use of pocket calculators.

Here’s a perfectly respectable way that someone in the 1880s could have computed $\sqrt{4213}$ to reasonably high precision. Let’s write

$x = \sqrt{4213}$.

Take the base-10 logarithm of both sides.

$\log_{10} x = \log_{10} \sqrt{4213} = \log_{10} (4213)^{1/2} = \displaystyle \frac{1}{2} \log_{10} 4213$.

Then log tables can be used to compute $\log_{10} 4213$.

Step 1. In our case, we’re trying to find $\log_{10} 4213$. We know that $\log_{10} 1000 = 3$ and $\log_{10} 10,000 = 4$, so the answer must be between $3$ and $4$. More precisely,

$\log_{10} 4213 = \log_{10} (1000 \times 4.213) = \log_{10} 1000 + \log_{10} 4.213 = 3 + \log_{10} 4.213$.

To find $\log_{10} 4.213$, we see from the table that

$\log_{10} 4.21 \approx 0.6243$ and $\log_{10} 4.22 = 0.6253$

So, to estimate $\log_{10} 4.213$, we will employ linear interpolation. That’s a fancy way of saying “Find the line connecting $(4.21,0.6243)$ and $(4.22,0.6253)$, and find the point on the line whose $x-$coordinate is $4.213$. Finding this line is a straightforward exercise in the point-slope form of a line:

$m = \displaystyle \frac{0.6253-0.6243}{4.22-4.21} = 0.1$

$y - 0.6243 = 0.1 (x - 4.21)$

$y = 0.6243 + 0.1 (4.213-4.21)$

$y = 0.6243 + 0.1(0.003) = 0.6246$

So we estimate $\log_{10} 4.213 \approx 0.6246$. Thus, so far in the calculation, we have

$\log_{10} \sqrt{4213} \approx \displaystyle \frac{1}{2} (3 + 0.6246) = 1.8123$

Step 2. We then take the antilogarithm of both sides. The term antilogarithm isn’t used much anymore, but the principle is still taught in schools: take $10$ to the power of both the left- and right-hand sides. We obtain

$\sqrt{4213} \approx 10^{1.8123} = 10^{1 + 0.8123} = 10^1 \times 10^{0.8123}$

The first part of the right-hand side is easy: $10^1 = 10$. For the second-part, we use the log table again, but in reverse. We try to find the numbers that are closest to $0.8123$ in the body of the table. In our case, we find that

$\log_{10} 6.49 = 0.8122$ and $\log_{10} 6.50 = 0.8129$.

Once again, we use linear interpolation to find the line connecting $(6.49,0.8122)$ and $(6.50,0.8129)$, except this time the $y-$coordinate of $0.8123$ is known and the $x-$coordinate is unknown.

$m = \displaystyle \frac{0.8129-0.8122}{6.50-6.49} = 0.07$

$y - 0.8122 = 0.07 (x - 6.49)$

$0.8123 - 0.8122 = 0.07 (x - 6.49)$

$x = 6.49 + \displaystyle \frac{0.0001}{0.07} = 6.4914\dots$

Since the table is only accurate to four significant digits, we estimate that $10^{0.8123} \approx 6.491$. Therefore,

$\sqrt{4213} \approx 10^1 \times 10^{0.8123} = 10 \times 6.491 = 64.91$

By way of comparison, the answer is $\sqrt{4213} \approx 64.9076\dots \approx 64.91$, rounding at the hundredths digit. Not bad, for a generation born before the advent of calculators.

With a little practice, one can do the above calculations with relative ease. Also, many log tables of the past had a column called “proportional parts” that essentially replaced the step of linear interpolation, thus speeding the use of the table considerably.

Log tables can be used for calculations more complex than finding a square root. For example, suppose I need to calculate

$x = \displaystyle \frac{(34.5)^3}{(912)^{2/5}}$

Using the log table, and without using a calculator, I find that

$\log_{10} x = 3 \log_{10} 34.5 - \displaystyle \frac{2}{5} \log_{10} 912$

$\log_{10} x = 3(1.5378) - \displaystyle \frac{2}{5} (2.9600)$

$\log_{10} x = 3.4294$

$x = 10^3 \cdot 10^{0.4294} = 1000 \cdot 2.688 = 2688$

That’s the correct answer to four significant digits. Using a calculator, we find the answer is $2688.186\dots$

# Square roots and logarithms without a calculator (Part 5)

I’m in the middle of a series of posts concerning the elementary operation of computing a square root. This is such an elementary operation because nearly every calculator has a $\sqrt{~~}$ button, and so students today are accustomed to quickly getting an answer without giving much thought to (1) what the answer means or (2) what magic the calculator uses to find square roots. I like to show my future secondary teachers a brief history on this topic… partially to deepen their knowledge about what they likely think is a simple concept, but also to give them a little appreciation for their elders.

One way that square roots can be computed without a calculator is by using log tables. This was a common computational device before pocket scientific calculators were commonly affordable… say, the 1920s.

As many readers may be unfamiliar with this blast from the past, Parts 3 and 4 of this series discussed the mechanics of how to use a log table. In Part 6, I’ll discuss how square roots (and other operations) can be computed with using log tables.

In this post, I consider the modern pedagogical usefulness of log tables, even if logarithms can be computed more easily with scientific calculators.

A personal story: In either 1981 or 1982, my parents bought me my first scientific calculator. It was a thing of beauty… maybe about 25% larger than today’s TI-83s, with an LED screen that tilted upward. When it calculated something like $\log_{10} 4213$, the screen would go blank for a couple of seconds as it struggled to calculate the answer. I’m surprised that smoke didn’t come out of both sides as it struggled. It must have cost my parents a small fortune, maybe over $1000 after adjusting for inflation. Naturally, being an irresponsible kid in the early 1980s, it didn’t last but a couple of years. (It’s a wonder that my parents didn’t kill me when I broke it.) So I imagine that requiring all students to use log tables fell out of favor at some point during the 1980s, as technology improved and the prices of scientific calculators became more reasonable. I regularly teach the use of log tables to senior math majors who aspire to become secondary math teachers. These students who have taken three semesters of calculus, linear algebra, and several courses emphasizing rigorous theorem proving. In other words, they’re no dummies. But when I show this blast from the past to them, they often find the use of a log table to be absolutely mystifying, even though it relies on principles — the laws of logarithms and the point-slope form of a line — that they think they’ve mastered. So why do really smart students, who after all are math majors about to graduate from college, struggle with mastering log tables, a concept that was expected of 15- and 16-year-olds a generation ago? I personally think that a lot of their struggles come from the fact that they don’t really know logarithms in the way that students of previous generation had to know them in order to survive precalculus. For today’s students, a logarithm is computed so easily that, when my math majors were in high school, they were not expected to really think about its meaning. For example, it’s no longer automatic for today’s math majors to realize that $\log_{10} 4213$ has to be between $3$ and $4$ someplace. They’ll just punch the numbers in the calculators to get an answer, and the process happens so quickly that the answer loses its meaning. They know by heart that $\log_{b} xy = \log_{b} x + \log_{b} y$ and that $\log_b b^x = x$. But it doesn’t reflexively occur to them that these laws can be used to rewrite $\log_{10} 4213$ as $3 + \log_{10} 4.213$. When encountering $10^{1.8123}$, their first thought is to plug into a calculator to get the answer, not to reflect and realize that the answer, whatever it is, has to be between $10$ and $100$ someplace. Today’s math majors can be taught these approximation principles, of course, but there’s unfortunately no reason to expect that they received the same training with logarithms that students received a generation ago. So none of this discussion should be considered as criticism of today’s math majors; it’s merely an observation about the training that they received as younger students versus the training that previous generations received. So, do I think that all students today should exclusively learn how to use log tables? Absolutely not.If college students who have received excellent mathematical training can be daunted by log tables, you can imagine how the high school students of generations past must have felt — especially the high school students who were not particularly predisposed to math in the first place. People like me that made it through the math education system of the 1980s (and before) received great insight into the meaning of logarithms. However, a lot of students back then found these tables as mystifying as today’s college students, and perhaps they did not survive the system because they found the use of the table to be exceedingly complex. In other words, while they were necessary for an era that pre-dated pocket calculators, log tables (and trig tables) were an unfortunate conceptual roadblock to a lot of students who might have had a chance at majoring in a STEM field. By contrast, logarithms are found easily today so that the steps above are not a hindrance to today’s students. That said, I do argue that there is pedagogical value (as well as historical value) in showing students how to use log tables, even though calculators can accomplish this task much quicker. In other words, I wouldn’t expect students to master the art of performing the above steps to compute logarithms on the homework assignments and exams. But if they can’t perform the above steps, then there’s room for their knowledge of logarithms to grow. And it will hopefully give today’s students a little more respect for their elders. # Square roots and logarithms without a calculator (Part 4) I’m in the middle of a series of posts concerning the elementary operation of computing a square root. In Part 3 of this series, I discussed how previous generations computed logarithms without a calculator by using log tables. In this post, I’ll discuss how previous generations computed, using the language of the time, antilogarithms. In Part 5, I’ll discuss my opinions about the pedagogical usefulness of log tables, even if logarithms can be computed more easily with scientific calculators. And in Part 6, I’ll return to square roots — specifically, how log tables can be used to find square roots. Let’s again go back to a time before the advent of pocket calculators… say, 1943. The following log tables come from one of my prized possessions: College Mathematics, by Kaj L. Nielsen (Barnes & Noble, New York, 1958). How to use the table, Part 5. The table can also be used to work backwards and find an antilogarithm. The term antilogarithm isn’t used much anymore, but the principle is still used in teaching students today. Suppose we wish to solve $\log_{10} x = 0.9509$, or $x = 10^{0.9509}$. Looking through the body of the table, we see that $9509$ appears on the row marked $89$ and the column marked $3$. Therefore,$10^{0.9509} \approx 8.93$. Again, this matches (to three and almost four significant digits) the result of a modern calculator. How to use the table, Part 6. Linear interpolation can also be used to find antilogarithms. Suppose we’re trying to evaluate $10^{0.9387}$, or find the value of $x$ so that$\log_{10} x = 0.9387$. From the table, we can trap $9387$ between $\log_{10} 8.68 \approx 0.9385$ and $\log_{10} 8.69 \approx 0.9390$ So we again use linear interpolation, except this time the value of $y$ is known and the value of $x$ is unknown: $m = \displaystyle \frac{0.9390-0.9385}{8.69-8.68} = 0.05$ $y - 0.9385= 0.05 (x - 8.68)$ $0.9387 - 0.9385= 0.05 (x - 8.68)$ $0.0002 = 0.05 (x-8.68)$ $0.004 =x-8.68$ $8.684 =x$ So we estimate $10^{0.9387} \approx 8.684$ This matches the result of a modern calculator to four significant digits: How to use the table, Part 7. How to use the table, Part 8. Note: Sorry, but I’m not sure what happened… when the post came up this morning (August 4), I saw my work in Parts 7 and 8 had disappeared. Maybe one of these days I’ll restore this. # Square roots and logarithms without a calculator (Part 3) I’m in the middle of a series of posts concerning the elementary operation of computing a square root. This is such an elementary operation because nearly every calculator has a $\sqrt{~~}$ button, and so students today are accustomed to quickly getting an answer without giving much thought to (1) what the answer means or (2) what magic the calculator uses to find square roots. I like to show my future secondary teachers a brief history on this topic… partially to deepen their knowledge about what they likely think is a simple concept, but also to give them a little appreciation for their elders. Today’s topic is the use of log tables. I’m guessing that many readers have either forgotten how to use a log table or else were never even taught how to use them. After showing how log tables were used in the past, I’ll conclude with some thoughts about its effectiveness for teaching students logarithms for the first time. This will be a fairly long post about log tables. In the next post, I’ll discuss how log tables can be used to compute square roots. To begin, let’s again go back to a time before the advent of pocket calculators… say, 1912. Before the advent of pocket calculators, most professional scientists and engineers had mathematical tables for keeping the values of logarithms, trigonometric functions, and the like. The following images come from one of my prized possessions: College Mathematics, by Kaj L. Nielsen (Barnes & Noble, New York, 1958). Some saint gave this book to me as a child in the late 1970s; trust me, it was well-worn by the time I actually got to college. With the advent of cheap pocket calculators, mathematical tables are a relic of the past. The only place that any kind of mathematical table common appears in modern use are in statistics textbooks for providing areas and critical values of the normal distribution, the Student $t$ distribution, and the like. That said, mathematical tables are not a relic of the remote past. When I was learning logarithms and trigonometric functions at school in the early 1980s — one generation ago — I distinctly remember that my school textbook had these tables in the back of the book. And it’s my firm opinion that, as an exercise in history, log tables can still be used today to deepen students’ facility with logarithms. In this post and Part 4 of this series, I discuss how the log table can be used to compute logarithms and (using the language of past generations) antilogarithms without a calculator. In Part 5, I’ll discuss my opinions about the pedagogical usefulness of log tables, even if logarithms can be computed more easily nowadays with scientific calculators. In Part 6, I’ll return to square roots — specifically, how log tables can be used to find square roots. How to use the table, Part 1. How do you read this table? The left-most column shows the ones digit and the tenths digit, while the top row shows the hundredths digit. So, for example, the bottom row shows ten different base-10 logarithms: $\log_{10} 9.90 \approx 0.9956, \log_{10} 9.91 \approx 0.9961, \log_{10} 9.92 \approx 0.9965,$ $\log_{10} 9.93 \approx 0.9969, \log_{10} 9.94 \approx 0.9974, \log_{10} 9.95 \approx 0.9978,$ $\log_{10} 9.96 \approx 0.9983, \log_{10} 9.97 \approx 0.9987, \log_{10} 9.98 \approx 0.9991,$ $\log_{10} 9.99 \approx 0.9996$ So, rather than punching numbers into a calculator, the table was used to find these logarithms. You’ll notice that these values match, to four decimal places, the values found on a modern calculator. How to use the table, Part 2. What if we’re trying to take the logarithm of a number between $1$ and $10$ which has more than two digits after the decimal point, like $\log_{10} 5.1264$? From the table, we know that the value has to lie between $\log_{10} 5.12 \approx 0.7093$ and $\log_{10} 5.13 \approx 0.7101$ So, to estimate $\log_{10} 5.1264$, we will employ linear interpolation. That’s a fancy way of saying “Find the line connecting $(5.12,0.7093)$ and $(5.13,0.7101)$, and find the point on the line whose $x-$coordinate is $5.1264$. The graph of$y = \log_{10} x\$ is not a straight line, of course, but hopefully this linear interpolation will be reasonably close to the correct answer.

Finding this line is a straightforward exercise in the point-slope form of a line:

$m = \displaystyle \frac{0.7101-0.7093}{5.13-5.12} = 0.08$

$y - 0.7093= 0.08 (x - 5.12)$

$y = 0.7093+ 0.08 (5.1264-5.12)$

$y = 0.7093+ 0.08(0.0064) = 0.7093 + 0.000512 = 0.709812$

Remembering that this log table is only good to four significant digits, we estimate $\log_{10} 5.1264 \approx 0.7098$.

With a little practice, one can do the above calculations with relative ease. Also, many log tables of the past had a column called “proportional parts” that essentially replaced the step of linear interpolation, thus speeding the use of the table considerably.

Again, this matches the result of a modern calculator to four decimal places:

How to use the table, Part 3. So far, we’ve discussed taking the logarithms of numbers between $1$ and $10$ and the antilogarithms of numbers between $0$ and $1$. Let’s now consider what happens if we pick a number outside of these intervals.

To find $\log_{10} 12,345$, we observe that

$\log_{10}12345 = \log_{10} (10,000 \times 1.2345)$

$\log_{10} 12345 = \log_{10} 10,000 + \log_{10} 1.2345$

$\log_{10} 12345 = 4 + \log_{10} 1.2345$

More intuitively, we know that the answer must lie between $\log_{10} 10,000 = 4$ and $100,000 = 5$, so the answer must be $4.\hbox{something}$. The value of $\log_{10} 1.2345$ is the necessary $\hbox{something}$.

We then find $\log_{10} 1.2345$ by linear interpolation. From the table, we see that

$\log_{10} 1.23 \approx 0.0899$ and $\log_{10} 1.24 \approx 0.0934$

Employing linear interpolation, we find

$m = \displaystyle \frac{0.0934-0.0899}{1.24-1.23} = 0.35$

$y - 0.0899= 0.35 (x - 1.23)$

$y = 0.0899+ 0.35 (1.2345-1.23)$

$y = 0.0899+ 0.35(0.0045) = 0.0899 + 0.001575 = 0.091475$

Remembering that this log table is only good to four significant digits, we estimate $\log_{10} 1.2345 \approx 0.0915$, so that $\log_{10} 12,345 \approx 4.0915$.

Again, this matches the result of a modern calculator to four decimal places (in this case, five significant digits):

How to use the table, Part 4.  Let’s now consider what happens if we pick a positive number less than $1$. To find $\log_{10} 0.00012345$, we observe that

$\log_{10}0.00012345 = \log_{10} \left( 1.2345 \times 10^{-4} \right)$

$\log_{10}0.00012345 = \log_{10} 1.2345 + \log_{10} 10^{-4}$

$\log_{10} 0.00012345 = -4 + \log_{10} 1.2345$

We have already found $\log_{10} 1.2345 \approx 0.0915$ by linear interpolation. We therefore conclude that $\log_{10} 12,345 \approx 0.0915 - 4 = -3.9085$. Again, this matches the result of a modern calculator to four decimal places (in this case, five significant digits):

So that’s how to compute logarithms without a calculator: we rely on somebody else’s hard work to compute these logarithms (which were found in the back of every precalculus textbook a generation ago), and we make clever use of the laws of logarithms and linear interpolation.

Log tables are of course subject to roundoff errors. (For that matter, so are pocket calculators, but the roundoff happens so deep in the decimal expansion — the 12th or 13th digit — that students hardly ever notice the roundoff error and thus can develop the unfortunate habit of thinking that the result of a calculator is always exactly correct.)

For a two-page table found in a student’s textbook, the results were typically accurate to four significant digits. Professional engineers and scientists, however, needed more accuracy than that, and so they had entire books of tables. A table showing 5 places of accuracy would require about 20 printed pages, while a table showing 6 places of accuracy requires about 200 printed pages. Indeed, if you go to the old and dusty books of any decent university library, you should be able to find these old books of mathematical tables.

In other words, that’s how the Brooklyn Bridge got built in an era before pocket calculators.

At this point you may be asking, “OK, I don’t need to use a calculator to use a log table. But let’s back up a step. How were the values in the log table computed without a calculator?” That’s a perfectly reasonable question, but this post is getting long enough as it is. Perhaps I’ll address this issue in a future post.