In 1981 three researchers at the University of Massachusetts, John Clement, Jack Lochhead and George Monk, set first-year engineering students a simple problem:

Write an equation using the variables $S$ and $P$ to represent the following statement: “There are six times as many students as professors at this university.” Use $S$ for the number of students and $P$ for the number of professors.

About 37% got it wrong. And almost all of those who got it wrong wrote the same thing:

$$6S=P.$$

Why 6S = P looks right

The equation $6S=P$ follows the order of the words: “six times… students… professors”. It is a word-by-word translation, like translating a text without understanding the sentence. Behind it lies a reading of the letter as a label: $S$ means “students”, just as in $6\,\text{m}$ the m means “metres”. Read that way, $6S=P$ says “six students for each professor”, which is exactly the right idea.

The problem is that in the equation $S$ is not the word “students”: it is the number of students. With $P=100$ professors there are $600$ students, and the correct equation must give $S=600$:

$$S=6P.$$

To check, just substitute numbers: with $P=100$, the wrong equation would give $6S=100$, that is, fewer than $17$ students. The letter as a label is a natural reading, which comes from everyday language and from units of measurement, and it has to be unlearned.

Küchemann’s six readings

In the same year the English researcher Dietmar Küchemann published the results of a large study of pupils aged 13 to 15, the CSMS project. He showed that pupils give letters at least six different interpretations, from the simplest to the most advanced:

Reading of the letter What the pupil does Sample task
Letter evaluated gives the letter a value straight away if $a+5=8$, what is $a$?
Letter not used solves without looking at it if $e+f=8$, what is $e+f+g$?
Letter as object treats the letter as a thing $2a+5b+a$ “is 3 apples and 5 bananas”
Specific unknown accepts that the letter is an unknown number if $e+f=8$, then $e+f+g=8+g$
Generalised number accepts that the letter can be several numbers if $c+d=10$ and $c
Variable relates how two letters change together which is larger, $2n$ or $n+2$?

The first three readings allow pupils to get many questions right without understanding any algebra. Only from the fourth on does the letter become a number. Küchemann found that most 13-year-olds did not get that far.

Fruit-salad algebra

Many teachers teach collecting like terms with an image: “$3a+2b+a$ is $3$ apples, $2$ bananas and $1$ more apple, so $4a+2b$.” The answer comes out right, but the image reinforces exactly the reading that causes $6S=P$: the letter as an abbreviation for a thing. With the same image, $a\cdot b$ means nothing (what is an apple times a banana?), and $a=2b$ seems to say “one apple is two bananas”. Researchers call this strategy fruit-salad algebra and advise against it.

Other traps

  • Joining digits. If $x=3$, what is $2x$? Some pupils answer $23$. In arithmetic, writing one digit next to another makes a number; in algebra, writing a number next to a letter multiplies. It is the same ambiguity of the blank space discussed in chapter 2.
  • Different letters, different numbers. Faced with $x+y=10$, many pupils think $x$ and $y$ cannot be equal, because “if they were equal they would have the same letter”. But $x=y=5$ is a perfectly valid solution.
  • A letter is always just one thing. Those who learned that $x$ is “the number being sought” find $y=2x+1$ strange, since its $x$ has no single value, and stranger still $f(x)$, in which $x$ is a dummy letter.
Exercise: which is larger, 2n or n + 2?

It depends on $n$. The two are equal when $2n=n+2$, that is, $n=2$. For $n>2$, $2n$ is larger (with $n=5$: $10$ against $7$). For $n<2$, $n+2$ is larger (with $n=0$: $0$ against $2$). Whoever answers “$2n$, because multiplying makes things bigger” is treating $n$ as a particular large number, and not as a variable.

Exercise: translate without inverting

“In a car park there are four times as many cars as motorbikes.” Use $C$ and $M$ for the numbers of cars and of motorbikes. The correct equation is $C=4M$. Test with numbers: if there are $10$ motorbikes, there are $40$ cars, and $40=4\cdot10$. The word-by-word translation, $4C=M$, would give $10$ motorbikes for $2.5$ cars, the opposite of the statement.

Chapter index · ← Dummy letters: the k of the sum and the t of the integral · How to read a letter: a guide and a glossary →