Ask a class what $(a+b)^2$ is and some will answer $a^2 + b^2$. Ask what $\sqrt{9 + 16}$ is and someone will say $3 + 4 = 7$. These errors are so common, and so similar from one school to another, that they cannot be chance. They follow a rule.

The rule that stretches too far

In 1982 the American researcher Marilyn Matz studied the algebra errors of high-school students and proposed that many of them come from false rules (malrules) built out of true ones. The most productive of all is the distributive law:

$$a(b + c) = ab + ac.$$

It is true, it appears early and it is used all the time. The student learns that “what is outside goes into each term” and, with nobody having said otherwise, extends the gesture to everything that looks like an operation applied to a sum:

$$(a+b)^2 \overset{?}{=} a^2 + b^2, \quad \sqrt{a+b} \overset{?}{=} \sqrt{a} + \sqrt{b}, \quad \frac{1}{a+b} \overset{?}{=} \frac{1}{a} + \frac{1}{b}, \quad \sin(a+b) \overset{?}{=} \sin a + \sin b.$$

All of them are false. All of them have the same shape as the distributive law. From the point of view of notation, the student is reading the writing very well: seeing a visual pattern and applying it consistently. What is missing is knowing that shape is not enough: the distributive law holds for multiplication because of what multiplication is, not because of how the brackets look.

The counterexample as a magnifying glass

The tool for dismantling a false rule is the counterexample: a single case in which it fails. With $a = 3$ and $b = 4$, $(a+b)^2 = 49$ and $a^2 + b^2 = 25$. That’s it: the rule is dead. A true rule, such as $(ab)^2 = a^2 b^2$, survives every test; the figure above lets you try.

Testing with small numbers is a reading habit, not only a checking one. Whoever writes a new equality and substitutes $a = 1$, $b = 2$ before going on catches most of their own errors. But the test has a limit, which chapter 4 showed: examples refute a rule but do not prove it.

The cancellation that works by chance

There is a famous error that sometimes gets it right:

$$\frac{16}{64} = \frac{1\not{6}}{\not{6}4} = \frac{1}{4}.$$

“Striking out” the $6$ on top with the $6$ below makes no sense at all: the $6$ of $16$ is worth six units and the one of $64$ is worth sixty. But the result is right! With two digits there are only four such fractions (not counting trivial ones like $\frac{11}{11}$): $\frac{16}{64}$, $\frac{19}{95}$, $\frac{26}{65}$ and $\frac{49}{98}$. They are rare enough to be curiosities, and dangerous enough to confirm, now and then, a false rule.

Exercise: when the dream is true

For which real numbers $a$ and $b$ does $(a+b)^2 = a^2 + b^2$ hold?

Expanding, $(a+b)^2 = a^2 + 2ab + b^2$. The equality requires $2ab = 0$, that is, $a = 0$ or $b = 0$. The false rule is right exactly when one of the terms is zero, which is why it sometimes survives badly chosen examples.

Exercise: the cancellation of 26/65

Check that $\frac{26}{65} = \frac{2}{5}$ and explain why “striking out the 6” is not a rule.

$\frac{26}{65}$: dividing by $13$ gives $\frac{2}{5}$. Correct. But $\frac{36}{63}$, striking out the $6$ in the same way, would give $\frac{3}{3} = 1$, while $\frac{36}{63} = \frac{4}{7}$. The coincidence of $\frac{26}{65}$ comes from the equation $\frac{10a + b}{10b + c} = \frac{a}{c}$, which has only a few solutions in digits; not from any property of repeated digits.

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