Here's part of "Tricky Living," copyright by Russ Walter, second edition. For newer info, read the 33rd edition of the "Secret Guide to Computers & Tricky Living" at


In my former life — before I tried to be a writer or a computer guy — I was a mathematician.


Torture your friends by giving them these puzzles about arithmetic.

Apples If you have 5 apples and eat all but 3, how many are left? Kids are tempted to say “2,” but the correct answer is 3.

Birds If you have 10 birds in a tree and shoot 1, how many are left in the tree? Kids are tempted to say “9,” but the correct answer is 0.

Corners If you have a 4-sided table and chop off 1 of the corners, how many corners are left on the table? Kids are tempted to say “3,” but the correct answer is 5.

Eggs Carl Sandberg, in his poem Arithmetic, asks this question:

If you ask your mother for one fried egg for breakfast, but she gives you two fried eggs and you eat both of them, who’s better in arithmetic: you or your mother?

Missing dollar Now that you’ve mastered the easy puzzles, try this harder one:

On a nice day in the 1940’s, three girls go into a hotel and ask for a triple. The manager says sorry, no triples are available, so he puts them in three singles, at $10 each. The girls go up to their rooms.

A few minutes later, a triple frees up, which costs just $25. So the manager, to be a nice guy, decides to move the girls into the triple and refund the $5 difference. He sends the bellboy up to tell the girls of their good fortune and move them into the triple.

While riding up in the elevator, the bellboy thinks to himself, “How can the girls split the $5? $5 doesn’t divide by 3 evenly. I’ll make it easier for them: I’ll give them just $3 — and keep $2 for myself.” So he gave the girls $3 and moved them into the triple.

Everybody was happy. The girls were happy to get refunds. The manager was happy to be a nice guy. And the bellboy was happy to keep $2.

Now here’s the problem: each girl spent $10 and got $1 back, so each girl spent $9. Altogether, the girls spent $9+$9+$9, which is $27, and the bellboy got $2. That makes $29. But we started with $30. What happened to the extra dollar?

Ask your friends that question and see how many crazy answers you get!

Here’s the correct answer:

At the end of the story, who has the $30?

The manager has $25, the bellboy has $2, and the girls have $3.

There’s no logical reason why taking the amount that the girls spent and adding it to what the bellboy received should give any meaningful number. But the nonsense total, $29, is close enough to $30 to be intriguing.

Here’s an alternative analysis:

The girls spent a net of $9+$9+$9, which is $27.

$25 of that went to the manager, and $2 went to the bellboy.

Coins Try this task:

Arrange 10 coins so they form 5 rows, each containing 4 coins.

“5 rows of 4 coins” would normally require a total of 20 coins, but if you arrange properly you can solve the puzzle. Hint: the rows must be straight but don’t have to be horizontal or vertical. Ask your friends that puzzle to drive them nuts.

Here’s the solution:

Draw a 5-pointed star. Put the coins at the 10 corners.

Which type are you?

Here’s Warren Buffet’s favorite saying about math.

There are 3 types of people: those who can count, and those who can’t.


Courses in statistics can be difficult. That’s why they’re called “sadistics.”

Lies Statisticians give misleading answers.

For example, suppose you’ve paid one person a salary of $1000, another person a salary of $100, another person a salary of $10, and two other people a salary of $1 each. What’s the “typical” salary you paid? If you ask that question to three different statisticians, they’ll give you three different answers!

One statistician will claim that the “typical” salary is $1, because it’s the most popular salary: more people received $1 than any other amount. Another statistician will claim that the “typical” salary is $10, because it’s the middle salary: as many people were paid more than $10 as were paid less. The third statistician will claim that the “typical” salary is $222.40, because it’s the average: it’s the sum of all the salaries divided by the number of people.

Which statistician is right? According to the Association for Defending Statisticians (started by my friends), the three statisticians are all right! The most common salary ($1) is called the mode; the middle salary ($10) is called the median; the average salary ($222.40) is called the mean.

But which is the “typical” salary, really? Is it the mode ($1), the median ($10), or the mean ($222.40)? That’s up to you!

If you leave the decision up to the statistician, the statistician’s answer will depend on who hired him.

If the topic is a wage dispute between labor and management, a statistician paid by the laborers will claim that the typical salary is low (just $1); a statistician paid by the management will claim that the typical salary is high ($222.40); and a statistician paid by the arbitrator will claim that the typical salary is reasonable ($10).

Which statistician is telling the whole truth? None of them!

A century ago, Benjamin Disraeli, England’s prime minister, summarized the whole situation in one sentence. He said:

There are 3 kinds of lies:

lies, damned lies, and statistics.


A course in “logic” is a blend of math and philosophy. It can be lots of fun — and also help you become a lawyer.

Beating your wife There’s the old logic question about how to answer this question:

Have you stopped beating your wife?

Regardless of whether you answer that question by saying “yes” or “no,” you’re implying that you did indeed beat your wife in the past.

Interesting number Some numbers are interesting. For example, some people think 128 is interesting because it’s “2 times 2 times 2 times 2 times 2 times 2 times 2.” Here’s a proof that all numbers are interesting:

Suppose some numbers are not interesting. For example, suppose 17 is the first number that’s not interesting. Then people would say, “Hey, that’s interesting! 17 has the very interesting property of being the first boring number!” But then 17 has become interesting! So you can’t have a first “boring” number, and all numbers are interesting!

Surprise test When I took a logic course at Dartmouth College, the professor began by warning me and my classmates:

I’ll give a surprise test sometime during the semester.

Then he told the class to analyze that sentence and try to deduce when the surprise test would be.

He pointed out that the test can’t be on the semester’s last day — because if the test didn’t happen before then, the students would be expecting the test when they walk into class on that last day, and it wouldn’t be a surprise anymore. So cross “the semester’s last day” off the list of possibilities.

Then he continued his argument:

But once you cross “the semester’s last day” off the list of possibilities, you realize the surprise test can’t be “the day before the semester’s last day” either, because the test would be expected then (since the test hadn’t happened already and couldn’t happen on the semester’s last day). Since the test would be expected then, it wouldn’t be a surprise. So cross “the day before last” off the list of possibilities.

Continuing in that fashion, he said, more and more days would be crossed off, until eventually all days would be crossed off the list of possibilities, meaning there couldn’t be a surprise test.

Then he continued:

But I assure you, there will be a test, and it will be a surprise when it comes.

Think about it.

Mathematicians versus engineers

The typical mathematician finds abstract concepts beautiful, and doesn’t care whether they have any “practical” applications. The typical engineer is exactly the opposite: the engineer cares just about practical applications.

Engineers complain that mathematicians are ivory-tower daydreamers who are divorced from reality. Mathematicians complain that engineers are too worldly and also too stupid to appreciate the higher beauties of the mathematical arts.

To illustrate those differences, mathematicians tell 3 tales.…

Boil water Suppose you’re in a room that has a sink, stove, table, and chair. A kettle is on the table. Problem: boil some water.

An engineer would carry the kettle from the table to the sink, fill the kettle with water, put the kettle onto the stove, and wait for the water to boil. So would a mathematician.

But suppose you change the problem, so the kettle’s on the chair instead of the table. The engineer would carry the kettle from the chair to the sink, fill the kettle with water, put the kettle onto the stove, and wait for the water to boil. But the mathematician would not! Instead, the mathematician would carry the kettle from the chair to the table, yell “now the problem’s been reduced to the previous problem,” and walk away.

Analyze tennis Suppose 1024 people are in a tennis tournament. The players are paired, to form 512 tennis matches; then the winners of those matches are paired against each other, to form 256 play-off matches; then the winners of the play-off matches are paired against each other, to form 128 further play-off matches; etc.; until finally just 2 players remain — the finalists — who play against each other to determine the 1 person who wins the entire tournament. Problem: compute how many matches are played in the entire tournament.

The layman would add 512+256+128+64+32+16+8+4+2+1, to arrive at the correct answer, 1023.

The engineer, too lazy to add all those numbers, would realize that the numbers 512, 256, etc., form a series whose sum can be obtained by a simple, magic formula! Just take the first number (512), double it, and then subtract 1, giving a final result of 1023!

But the true mathematician spurns the formula and searches instead for the problem’s underlying meaning. Suddenly it dawns on him! Since the problem said there are “1024 people” but just 1 final winner, the number of people who must be eliminated is “1024 minus 1,” which is 1023, and so there must be 1023 matches!

The mathematician’s calculation (1024-1) is faster than the engineer’s. But best of all, the mathematician’s reasoning applies to any tournament, even if the number of players isn’t a magical number such as 1024. No matter how many people play, just subtract 1 to get the number of matches!

Prime numbers Mathematicians are precise, physicists somewhat less so, chemists even less so. Engineers are even less precise and sometimes less intellectual. To illustrate that view, mathematicians tell the tale of prime numbers.

First, let me explain some math jargon. The counting numbers are 1, 2, 3, etc. A counting number is called composite if you can get it from multiplying a pair of other counting numbers. For example:

  6 is composite because you can get it from multiplying 2 by 3.

  9 is composite because you can get it from multiplying 3 by 3.

15 is composite because you can get it from multiplying 3 by 5.

A counting number that’s not composite is called prime. For example, 7 is prime because you can’t make 7 from multiplying a pair of other counting numbers. Whether 1 is “prime” depends on how you define “prime,” but for the purpose of this discussion let’s consider 1 to be prime.

Here’s how scientists would try to prove this theorem:

All odd numbers are prime.

Actually, that theorem is false! All odd numbers are not prime! For example, 9 is an odd number that’s not prime. But although 9 isn’t prime, the physicists, chemists, and engineers would still say the theorem is true.

The physicist would say, slowly and carefully:

1 is prime. 3 prime. 5 is prime. 7 is prime.

9? — no.

11 is prime. 13 is prime.

9 must be just experimental error, so we can ignore it. All odd numbers are prime.

The chemist would rush for results and say just this:

1 is prime, 3 is prime. 5 is prime. 7 is prime.

That’s enough evidence. All odd numbers are prime.

The engineer would be the crudest and stupidest of them all. He’d say the following as fast as possible (to meet the next deadline for building his rocket, which will accidentally blow up):

Sure, 1 is prime, 3 is prime, 5 is prime, 7 is prime, 9 is prime, 11 is prime, 13 is prime, 15 is prime, 17 is prime, 19 is prime, all odd numbers are prime!


Every few years, authors of math textbooks come out with new editions, to reflect the latest fads. Here’s an example, as reported (and elaborated on) by Reader’s Digest (in February 1996), Recreational & Educational Computing (issue #91), John Funk (and his daughter), ABC News Radio WTKS 1290 (in Savannah), and others:

Teaching math in 1960: traditional math

A logger sells a truckload of lumber for $100.

His cost of production is 4/5 of the price.

What’s his profit?

Teaching math in 1965: simplified math

A logger sells a truckload of lumber for $100.

His cost of production is 4/5 of the price, or $80.

What’s his profit?

Teaching math in 1970: new math

A logger exchanges a set L of lumber for a set M of money.

The cardinality of set M is 100. Each element is worth $1.

Make 100 dots representing the elements of M.

The set C (cost of production) contains 20 fewer points than set M.

Represent the set C as a subset of set M and answer this question:

what’s the cardinality of the set P of profits?

Teaching math in 1975: feminist-empowerment math

A logger sells a truckload of lumber for $100.

Her cost is $80, and her profit is $20.

Your assignment: underline the number 20.

Teaching math in 1980: environmentally conscious math

An unenlightened logger cuts down beautiful trees, desecrating the precious forest for $20. Write an essay explaining how you feel about that way to make money. How did the forest’s birds and squirrels feel?

Teaching math in 1985: computer-based math

A logger sells a truckload of lumber for $100.

His production costs are 80% of his revenue.

On your calculator, graph revenue versus costs.

On your computer, run the LOGGER program to determine the profit.

Teaching math in 1990: Wall Street math

By laying off 40% of its loggers, a company improves its stock price from $80 to $100. How much capital gain per share does the CEO make by exercising his options at $80? Assume capital gains have become untaxed to encourage investment.

Teaching math in 1995: managerial math

A company outsources all its loggers. The firm saves on benefits; and whenever demand for its products is down, the logging workforce can be cut back easily. The average logger employed by the company earned $50,000 and had a 3-week vacation, nice retirement plan, and medical insurance. The contracted logger charges $30 per hour. Based on that data, was outsourcing a good move? If a laid-off logger comes into the logging company’s corporate headquarters and goes postal, mowing down 16 executives and a couple of secretaries, was outsourcing the loggers still a good move?

Teaching math in 2000: tax-based math

A logger sells a truckload of lumber for $100. His cost of production is 4/5 of the price. After taxes, why did he bother?

Teaching math in 2005: profit-pumping math

A logger sells a truckload of lumber for $100.

His production cost is $120.

How did Arthur Anderson determine that his profit margin is $60?

Teaching math in 2010: multicultural math

Un maderero vende un camión de madera para $100. Su coste de producción es $80….

Winston Churchill

Winston Churchill (who was England’s prime minister) said:

I had a feeling once about Mathematics — that I saw it all. Depth beyond Depth was revealed to me: the Byss and the Abyss. I saw — as one might see the transit of Venus or even the Lord Mayor’s Show — a quantity passing through infinity and changing its sign from plus to minus. I saw exactly why it happened and why the tergiversation was inevitable — but it was after dinner and I let it go.

Terrorist mathematicians

A colleague passed me this e-mail, forwarded anonymously:

A teacher was arrested because he attempted to board a flight while possessing a ruler, protractor, and calculator. Attorney General Alberto Gonzales said he believes the man’s a member of the notorious Al-gebra movement. The man’s been charged with carrying weapons of math instruction.

“Al-gebra is a problem for us,” Gonzales said. “Its followers desire solutions by means and extremes and sometimes go off on tangents in search of absolute values. They use secret code names like ‘x’ and ‘y’ and refer to themselves as “unknowns,’ but we’ve determined they belong to a common denominator of the axis of medieval, with coordinates in every country.”

When asked to comment on the arrest, George W. Bush said, “If God had wanted us to have better weapons of math instruction, He’d have given us more fingers and toes.” Aides told reporters they couldn’t recall a more intelligent or profound statement by the President.


In math, the most famous constant is pi, which is roughly 3.14. But another famous math constant is 1089. It’s the favorite constant among math magicians because it creates this trick.…

Write down any three-digit number “whose first digit differs from the last digit by more than 1.” For example:

852 is okay, since its first digit (8) differs from the last digit (2) by 6,

which is more than 1.

479 is okay, since its first digit (4) differs from the last digit (9) by 5,

which is more than 1.

282 is not okay, since the difference between 2 and 2 is 0.

Take your three-digit number, and write it backwards. For example, if you picked 852, you now have on your paper:



You have two numbers on your paper. One is smaller than the other. Subtract the small one from the big one:




Take your answer, and write it backward:





Add the last two numbers you wrote:






Notice that the final answer is 1089.

1089 is the final answer, no matter what three-digit number you started with (if the first and last digits differ by more than 1).

Here’s another example:

Take a number:                          724

Write it backward & subtract: -427


Write it backward & add:        +792


Here’s another example:

Take a number:                          365

Write it backward & subtract: 563



Write it backward & add:        +891


Yes, you always get 1089!

Proof To prove you always get 1089, use algebra: make letters represent the digits, like this.…

                                 Hundreds     Tens              Ones

Take a number:          A         B         C

Write it backwards:    C         B         A

To subtract the bottom (C B A) from the top (A B C), the top must be bigger. So in the hundreds column, A must be bigger than C. Since A is bigger than C, you can’t subtract A from C in the ones column, so you must borrow from the B in the tens column, to produce this:

                                 Hundreds     Tens              Ones

                                 A         B-1       C+10

                                 C         B         A

Now you can subtract A from C+10:

                                 Hundreds     Tens              Ones

                                 A         B-1       C+10

                                  C         B         A


In the tens column, you can’t subtract B from B-1, so you must borrow from the A in the hundreds column, to produce this:

                                 Hundreds     Tens              Ones

                                 A-1       B-1+10    C+10

                                 C         B         A


Complete the calculation:

                                 Hundreds     Tens              Ones

Start with this:           A-1       B-1+10    C+10

Subtract this:             C         B         A

Get this result:           A-1-C     9         C+10-A

Write it backwards:    C+10-A    9         A-1-C

Get this total:            10        8         9

Rounded Rectangular Callout: 9, plus the 1 that was carried             


Don’t burn your arm I call 1089 the “don’t burn your arm” number, because of this trick suggested by Irving Adler in The Magic House of Numbers:

Tell a friend to write a 3-digit number whose first & last digits differ by more than 1. Tell him to write the number backwards, subtract, write that backwards, and add. Tell him to burn the paper he did the figuring on. Put your arm in the ashes. When you take your arm out, the number 1089 will be mysteriously written on your arm in black. (The way you get 1089 to appear is to write “1089” on your arm with wet soap before you begin the trick. When you put your arm in the ashes, the answer will stick to the soap.) The trick works — if you don’t burn your arm.

April Fools Irving Adler also suggested this trick:

Tell a friend to write a 3-digit number whose first & last digits differ by more than 1. Say to write the number backwards, subtract, write that backwards, and add. At this point, you know the friend has 1089, but don’t let on. Just continue, by giving him these directions:

multiply by a million

subtract 733361573

under each 3 in the answer, write an L

under each 6, write an F

under each 5, write an O

under each 8, write an I

under each 4, write an R

under each 2, write a P

under each 7, write an A

read it backwards


Variants That procedure (reverse then subtract, reverse then add) gives 1089 if you begin with an appropriate 3-digit number. If you begin with a 2-digit number instead, you get 99.

If you begin with a 4-digit number instead, you get 10989 or 10890 or 9999, depending on which of the 4 digits are the biggest. If you begin with a 5-digit number, you get 109989 or 109890 or 99099. Notice that the answers for 4-digit and 5-digit numbers — 10989, 10890, 9999, 109989, 109890, and 99099 — are all formed from the number 99 and 1089.

Pythagorean theorem

The most amazing math discovery made by Greeks is the Pythagorean theorem. It says that in a right triangle (a triangle including a 90° angle), a²+b²=c², where “c” is the length of the hypotenuse (the longest side) and “a” & “b” are the lengths of the legs (the other two sides). It says that in this diagram —















c’s square is exactly as big (has the same area) as a’s square and b’s square combined.

The Chinese discovered the same truth, perhaps earlier.

Why is the Pythagorean theorem true? How do you prove it?

You can prove it in many ways. The 2nd edition of a book called The Pythagorean Proposition contains many proofs (256 of them!), collected in 1940 by Elisha Scott Loomis when he was 87 years old. Here are the 5 most amazing proofs.…

3-gap proof Draw a square, where each side has length a+b. In each corner of that square, put a copy of the triangle you want to analyze, like this:










Now the square contains those 4 copied triangles, plus 1 huge gap in the middle. That gap is a square where each side has length c, so its area is c².

Now move the bottom 2 triangles up, so you get this:











The whole picture is still “a square where each side has length a+b,” and you still have 4 triangles in it; but instead of a big gap whose area is c², you have two small gaps, of sizes a² and b². So c² is the same size as a²+b².

1-gap proof Draw the same picture that the 3-gap proof began with. You see the whole picture’s area is (a+b)². You can also see that the picture is cut into 4 triangles (each having an area of ab/2) plus the gap in the middle (whose area is c²). Since the whole picture’s area must equal the sum of its parts, you get:

(a+b)² = ab/2 + ab/2 + ab/2 + ab/2 + c²

In this proof, instead of “moving the bottom 2 triangles,” we use algebra. According to algebra’s rules, that equation’s left side becomes a² + 2ab + b², and the right side becomes 2ab + c², so the equation becomes:

a² + 2ab + b² = 2ab + c²

Subtracting 2ab from both sides of that equation, you’re left with:

a² + b² = c²

1-little-gap proof Draw a square, where each side has length c. In each corner of that square, put a copy of the triangle you want to analyze, like this:








The whole picture’s area is c². The picture is cut into 4 triangles (each having an area of ab/2) plus the little gap in the middle, whose area is (b-a)². Since the whole picture’s area must equal the sum of its parts, you get:

c² = ab/2 + ab/2 + ab/2 + ab/2 + (b-a)²

According to algebra’s rules, that equation’s right side becomes 2ab + (b² - 2ba + a²). Then the 2ab and the -2ba cancel each other, leaving you with a² + b², so the equation becomes:

c² = a² + b²

1-segment proof Draw the triangle you’re interested in, like this:




Unlike the earlier proofs, which make you draw many extra segments (short lines), this proof makes you draw just one extra segment! Make it perpendicular to the hypotenuse and go to the right angle:




The original big triangle (whose sides have lengths a, b, and c) has the same-size angles as the tiny triangle (whose sides have lengths x and a), so it’s “similar to” the tiny triangle, and so the big triangle’s ratio of “shortest side to hypotenuse” (a/c) is the same as the tiny triangle’s ratio of “shortest side to hypotenuse” (x/a). Write that equation:

a/c = x/a

Multiplying both sides of that equation by ac, you discover what a² is:

a² = xc

Using similar reasoning, you discover what b² is:

b² = yc

Adding those two equations together, you get:

a² + b² = (x+y)c

Since x+y is c, that equation becomes:

a² + b² = c²

1-segment general proof Draw the triangle you’re interested in, like this:




As in the previous proof, draw one extra segment, perpendicular to the hypotenuse and going to the right angle:




Now you have 3 triangles: the left one, the rightmost one, and the big one.

Since the left triangle’s area plus the rightmost triangle’s area equals the big triangle’s area, and since the 3 triangles are similar to each other (“stretched” versions of each other, as you can prove by looking at their angles), any area constructed from “parts of the left triangle” plus the area constructed from “corresponding parts of the rightmost triangle” equals the area constructed from “corresponding parts of the big triangle.” For example, the area constructed by drawing a square on the left triangle’s hypotenuse (a²) plus the area constructed by drawing a square on the rightmost triangle’s hypotenuse (b²) equals the area constructed by drawing a square on the big triangle’s hypotenuse (c²).

Which proof is the best? The 3-gap proof is the most visually appealing, but it bothers mathematicians who are too lazy to draw (construct) so many segments. (It also requires you to prove the gap is indeed a square, whose angles are right angles, but that’s easy.)

The 1-gap proof uses fewer lines by relying on algebra instead. It’s fine if you like algebra, awkward if you don’t. The 1-little-gap proof uses algebra slightly differently.

The 1-segment proof appeals to mathematicians because it requires constructing just 1 segment, but you can’t understand it until you’ve learned the laws of similar triangles. This proof was invented by Davis Legendre in 1858.

The 1-segment general proof is the most powerful because its thinking generalizes to any area created from the 3 triangles, not just square areas. In any right triangle:

The area of a square drawn on the hypotenuse (c²) is the sum of the areas of squares drawn on the legs (a² + b²).

The area of a circle drawn on the hypotenuse (using the hypotenuse as the diameter) is the sum of the areas of circles drawn on the legs.

The area of any blob (such as a square or circle or clown’s head) drawn on the hypotenuse is the sum of the areas of similarly-shaped blobs drawn on the legs.

That proof was invented by a 19-year-old kid (Stanley Jashemski in Youngstown, Ohio) in 1934.


To understand the concept of math ugliness, remember these math definitions:

The numbers 0, 1, 2, 3, etc., are called
whole numbers.

Those numbers and their negatives (-1, -2, -3, etc.) are all called integers.

The integers and fractions made from them (1/4, 2/3, -7/5, etc.) are all called rational numbers (because they’re all simple fractions, simple ratios).

All numbers on the number line are called
real numbers: they include all the rational numbers but also include irrational numbers (such as “pi” and “the square root of 2”), which can’t be expressed accurately as fractions made of integers.

Now you can tackle the 3 rules of ugliness:

1. Most things are ugly.

2. Most things you’ll see are nice.

3. Every ugly thing is almost nice.

More precisely:

Suppose you have a big set of numbers (such as the set of all real numbers), and you consider a certain subset of those numbers to be “nice” (such as the set of all rational numbers). The 3 rules of ugliness say:

1. Most members of the big set aren’t in the nice subset. (For example, most real numbers aren’t rational.)

2. When you operate on most members of the nice subset, you stay in the nice subset. (For example, if you add, subtract, multiply, or divide rational numbers, you get another rational number, if you don’t divide by 0.)

3. Ever member of the big set can be approximated by members of the nice subset. (For example, every irrational number can be approximated by rational numbers.)

In different branches of math, those same 3 rules keep cropping up, using different definitions of what’s “ugly” and “nice.”

The rules apply to people, too:

1. Most people aren’t like you. You’ll tend to think their behaviors are ugly.

2. Most people you’ll meet will appeal to you, because you’ll tend to move to a neighborhood or career composed of people like you.

3. The “ugly” people are actually almost like you: once you make an attempt to understand them, you’ll discover they really aren’t as different from you as you thought!

How math should be taught

I have complaints about how math is taught. Here’s a list of my main complaints. If you’re a mathematician, math teacher, or top math student, read the list and phone me at 603-666-6644 if you want to chat about details or hear about my other complaints, most of which result from research I did in the 1960’s and 1970’s. (On the other hand, if you don’t know about math and don’t care, skip these comments.)

Percentages Middle-school students should learn how to compute percentages (such as “What is 40% of 200?”); but advanced percentage questions (such as “80 is 40% of what?” and “80 is what percent of 200?”) should be delayed until after algebra, because the easiest way to solve an advanced percentage question is to turn the question into an algebraic equation by using these tricks:

change “what” to “x”

change “is” to “=”

change “percent” to “/100”

change “of” to “·”

Graphing a line To graph a line (such as “y = 5 + 2x”), students should be told to use this formula:

the graph of the equation y = h + sx

is a line whose height (above the origin) is h

and whose slope is s

So to graph y = 5 + 2x, put a dot that’s a distance of 5 above the origin; then draw a line that goes through that dot and has a slope of 2.

The formula “y = h + sx” is called the “hot sex” formula (since it includes h + sx). It’s easier to remember than the traditional formula, which has the wrong letters and wrong order and looks like this:

the graph of the equation y = mx + b

is a line whose height (above the origin) is b

and whose slope is m

Imaginary numbers Imaginary numbers (such as “i”) should be explained before the quadratic formula, so the quadratic formula can be stated simply (without having to say “if the determinant is non-negative”).

Factoring Students should be told that every quadratic expression (such as x² + 6x + 8) can be factored by this formula:

the factorization of x² + 2ax + c is

(x+a+d)(x+a-d), where d=Öa²-c

For example:

to factor x² + 6x + 8,

realize that a=3 and c=8,

so d=1 and the factorization is (x+3+1)(x+3-1),

which is (x+4)(x+2)

As you can see from that example, the a (which in the example is 3) is the average of the two final numbers (4 and 2). That’s why it’s called a.

The d (which is 1) is how much each final number differs from a (4 and 2 each differ from 3 by 1). That’s why it’s called d. You can call d the difference or divergence or displacement.

Here’s another reason why it’s called d: it’s the determinant, since it determines what kind of final answer you’ll get (rational, irrational, imaginary, or single-root). You can also call d the discriminant, since it lets you discriminate among different kinds of answers.

Quadratic equations To solve any quadratic equation (such as “x² + 6x + 8 = 0”), you can use that short factoring formula. For example:

to solve “x² + 6x + 8 = 0,”

factor it to get “(x+4)(x+2) = 0,”

whose solutions are -4 and -2

Another way to solve a quadratic equation is to use “Russ’s quadratic formula,” which is:

the solution of “x² = 2bx+c” is b ± Öb²+c

That’s much shorter and easier to remember than the traditional quadratic formula, though forcing an equation into the form “x2 = 2bx+c” can sometimes be challenging. Here’s an application:

to solve x²=6x+16,

realize that b=3 and c=16,

so the solution is 3±Ö25, which is 3±5,

which is 8 or -2

Prismoid formula Students should be told that the volume of any reasonable solid (such as a prism, cylinder, pyramid, cone, or sphere) can be computed from this prismoid formula:

volume =

height • (area of the typical cross-section)

where “area of the typical cross-section” means (top + bottom + 4 • middle)/6, where

“top” means “area of top cross-section”

“bottom” means “area of bottom cross-section”

“middle” means “area of halfway-up cross-section”

That formula can be written more briefly, like this:

V = H (T + B + 4M)/6,

where V means volume,

H means height,

T means top cross-section’s area

B means bottom cross-section’s area

M means middle cross-section’s area

For example, the volume of a pyramid (whose height is H and whose base area is L times W) is:

H (0 + LW + 4(L/2)(W/2))/6, which is

H (LW + 4LW/4)/6, which is

H (LW + LW)/6, which is

H (2LW)/6, which is


The volume of a cone (whose height is H and whose base area is πr²) is:

H (0 + πr² + 4π(r/2)²)/6, which is

H (πr² + 4πr²/4)/6, which is

H (πr² + πr²)/6, which is

H (2πr²)/6, which is

H πr²/3

The volume of a sphere (whose radius is r) is:

(2r) (0 + 0 + 4πr²)/6, which is

2r (4πr²)/6, which is


In the prismoid formula, V = H (T + B + 4M)/6, the “4” is the same “4” that appears in Simpson’s rule (which is used in calculus to find the area under a curve). The formula gives exactly the right answer for any 3-D shape whose sides are “smooth” (so you can express the cross-sectional areas as a quadratic or cubic function of the distance above the base). To prove the prismoid formula works for all such shapes, you must study calculus.

Balanced curriculum Math consists of many topics. Schools should reevaluate which topics are most important.

All students, before graduating from high school, should taste what statistics and calculus are about, since they’re used in many fields. For example, economists often talk about “marginal profit,” which is a concept from calculus. Students should also be exposed to other branches of math, such as matrices, logic, topology, and infinite numbers.

The explanation of Euclidean geometry should be abridged, to make room for other topics that are more important, such as coordinate geometry, which leads to calculus.

Like Shakespeare, Euclid’s work is a classic that should be shown to students so they can savor it and enjoy geometric examples of what “proofs” are; but after half a year of that, let high-school students move on to other topics that are more modern and more useful, to see examples of how proofs are used in other branches of math.

Too much time is spent analyzing triangles.

For example, consider the experience of John Kemeny, who headed Dartmouth College’s math department (and also invented the Basic programming language and later became Dartmouth College’s president). When he was a high-school student, his teacher told him to master “trigonometry, the study of analyzing triangles”; but for the next 20 years, he never had to analyze another triangle, even though he was a mathematician. That trigonometry course was totally useless!

Finally, one day, he bought a plot of land that was advertised as being “an acre, more or less.” He wanted to discover whether it was more or less, so he had survey it and analyze triangles. (The plot turned out to be more than an acre.)

When he told that tale to me and my classmates at Dartmouth, he then went on to make his point: mathematicians don’t have much use for analyzing triangles, though they do have use for how trigonometric functions (such as sine and cosine) help analyze circles (and circular motion and periodic motion). So let’s spend less time on triangles and more time on other topics!

Infinitesimals Students should be told about infinitesimals, because they make calculus easier to understand.

Specifically, there’s an infinitesimal number, called epsilon (or є or simply e), which is positive (greater than zero) but so tiny that its square is 0:

є² = 0

You might say “there’s no such number,” but we can invent it, just like mathematicians invented the “imaginary” number i whose square is -1. The invention of i simplified algebra, by making the quadratic formula more understandable. The invention of є simplifies calculus, by making derivatives more understandable.

To use є, construct the
extended real numbers, which consist of numbers of the form a + bє (where “a” and “b” are ordinary “real” numbers). Add and multiply extended real numbers as you’d expect (bearing in mind that є² is 0), like this:

(a + bє) + (c + dє) = (a+c) + (b+d)є

(a + bє) • (c + dє) = ac + (ad+bc)є

For example:

(9+12є) + (2+4є) = 11+16є

(9+12є) • (2+4є) = 18 + (36+24)є, which is 18+60є

You can define order:

“a+bє < c+dє” means “a<c or (a=c and b<d)”

Those definitions of addition, subtraction, multiplication, and order obey the traditional “rules of algebra” except for one rule: in traditional algebra, every non-zero number has a reciprocal (a number you can multiply it by to get 1), but unfortunately є has no reciprocal.

If x is an extended real number, it has the form a + bє, where a and b are each real. The a is called the real part of x. For example, the real part of 3 + 7є is 3.

A number is called infinitesimal if its real part is 0. For example, є and 2є are infinitesimal; so is 0.

Infinitesimals are useful because they let you define the “derivative” of f(x) easily, by computing f(x+є):

Define the differential of f(x), which is written d f(x), to mean f(x+є) - f(x). For example, dx² is (x+є)²-x², which is (x²+2xє+є²)-x², which is 2xє (since є²=0), which is 2x dx (since dx turns out to be є).

Define the derivative of f(x) to mean
(d f(x)) divided by є. For example, the derivative of x² is (2xє)/є, which is 2x. The definition of the derivative of f(x) can also be written as (d f(x))/dx, since dx is є.

Define the limit, as x approaches p, of f(x) to mean the real part of f(p+є). For example, the limit, as x approaches 0, of x/x is the real part of (0+є)/(0+є), which is the real part of є/є, which is the real part of 1, which is 1.

Define f(x) is continuous at p to mean:

for all b, f(p+bє) – f(p) is infinitesimal.

For example, the function “3 if x£=0, 4 if x>0” isn’t continuous at 0, since f(0+1є)-f(0) is 4-3, which is 1, which isn’t infinitesimal.

Define f(x) is differentiable at p to mean:

for all b, f(p+bє) = f(p) + b (the derivative of f(x) at p).

Then calculations & proofs about derivatives and limits become easy, especially when you define sin є to be є and define cos є to be 1.