What’s Wrong with This Proof That 1 = 2?

A proof in algebra is a chain of claims about numbers. Each line says that two numbers are equal. The letters stand for numbers we haven’t been told, but they are still numbers, so we can test the chain by choosing some.

Take a = b = 3 in your cousin’s proof:

Line	What it says	With a = b = 3	True?
1	a = b	3 = 3	yes
2	$a^2 = ab$	9 = 9	yes
3	$a^2 - b^2 = ab - b^2$	0 = 0	yes
4	(a + b)(a − b) = b(a − b)	6 × 0 = 3 × 0	yes
5	a + b = b	6 = 3	no
6	2b = b	6 = 3	no
7	2 = 1	2 = 1	no

The first four lines are true. Line 5 is false. So the damage was done between line 4 and line 5, and nowhere else: the lines after it are false because they inherited a false line, not because their own steps are wrong.

What happened between 4 and 5? Both sides were divided by a − b. And since a = b, a − b is 0. Line 4 says 6 × 0 = 3 × 0, which is true: both sides are 0. Dividing both sides by 0 to get 6 = 3 is the broken step.

Try any other number for a and b, as long as they are equal: 5, 100, −2. Line 4 is always some number times 0 equals another number times 0, and line 5 is always false. The trick is the same every time; the letters only hide it. (At a = b = 0 the proof divides by zero twice: line 5 comes out 0 = 0 by luck, and the next false line comes after line 6, dividing by b, which is 0 too.)

A test that catches mistakes, not a proof of correctness

Putting numbers in is powerful, but it only works one way. If a line comes out false, something before it is wrong. If every line comes out true, that doesn’t prove the steps were right. A wrong step can land on a true line by luck.

The famous example: cancel the 6s in $\frac{16}{64}$ and you get $\frac{1}{4}$, which is correct! But “crossing out a digit that appears on top and bottom” is not a rule of arithmetic: try it on $\frac{12}{24}$ and you get $\frac{1}{4}$ instead of $\frac{1}{2}$. A right answer reached by a wrong step is called a howler. Leave out dull cases like $\frac{30}{60}$, where two 0s are crossed out, and among fractions less than 1 with two-digit numbers on top and bottom only four work by luck this way: $\frac{16}{64}$, $\frac{19}{95}$, $\frac{26}{65}$ and $\frac{49}{98}$.

So which steps are good ones? Here is what algebra really allows. If two numbers are equal and you do exactly the same thing to both, they stay equal. You may:

• add the same number to both sides;
• take the same number from both sides;
• multiply both sides by the same number;
• divide both sides by the same number, as long as it is not 0.

The first three need no condition at all. The fourth needs one, and the reason is what division means.

Why you can’t divide by zero

12 ÷ 3 = 4 because 4 × 3 = 12: dividing asks which number, times 3, gives 12. So 12 ÷ 0 would ask which number, times 0, gives 12. Every number times 0 is 0, so no number works. And 0 ÷ 0 asks which number, times 0, gives 0, and every number works: 1, 7, a million. A question with no answer, or with every answer, can’t give one definite number. So division by zero is undefined: it isn’t a very big number or zero or anything else; it has no meaning, and a proof may not do it.

Canceling: the real rule

Dividing both sides by the same thing is usually called canceling. Written out properly, the rule is:

$$\text{if } c \times x = c \times y \text{ and } c \neq 0, \text{ then } x = y.$$

Mathematicians call this the cancellation law. It is easy to picture. Think of c × x as the area of a rectangle x long and c wide. Two rectangles of the same width and the same area must have the same length: that is all the rule says. Unless the width is 0.

For every a bigger than 3, the two rectangles have the same width and different lengths, so their areas differ, and line 4 is simply false. Only at a = 3 does line 4 become true, and only because both rectangles have flattened into lines with no area at all. Equal areas of 0 tell you nothing about the lengths: a line 6 long and a line 3 long both have area 0. Canceling the width there pretends they do.

Steps that only go one way

There is a second, quieter lesson in line 4. Multiplying both sides by 0 is allowed: it keeps true lines true. But it also turns false lines into true ones. Start from the false 6 = 3, multiply both sides by 0, and you get the true 0 = 0. Some steps can be walked back, and some can’t:

Step	Keeps true lines true?	Can be undone?
Add or take away a number	yes	yes: take it away or add it back
Multiply by a number that isn’t 0	yes	yes: divide by it
Multiply by 0	yes	no: everything becomes 0 = 0
Divide by 0	not a step at all	—
Square both sides	yes	no: $(-2)^2 = 2^2$, but −2 ≠ 2

A step that can be undone gives an equivalent equation: one with exactly the same solutions. A step that can’t be undone is still a fair step forwards, but the new line may be true in cases where the old one wasn’t. That is exactly what your cousin’s proof did, in reverse: from line 4, which holds only because a − b = 0, it jumped back across a multiplication by 0.

This is what solving an equation really means. Writing 2x + 1 = 7, then 2x = 6, then x = 3 says: if a number makes the first line true, it must be 3. Every step here can be undone, so 3 really works. When a step can’t be undone, check each answer in the first equation.

Your cousin’s proof is the most famous of its kind, but there are many others, and each hides a different broken rule. Here are three, harder each time. Use the widget above to check them: choose the proof under it.

2 = 3, by squares

$$\begin{aligned} 4 - 10 &= 9 - 15 \\ 4 - 10 + \tfrac{25}{4} &= 9 - 15 + \tfrac{25}{4} \\ \left(2 - \tfrac{5}{2}\right)^2 &= \left(3 - \tfrac{5}{2}\right)^2 \\ 2 - \tfrac{5}{2} &= 3 - \tfrac{5}{2} \\ 2 &= 3 \end{aligned}$$

The first line is true: both sides are −6. Adding $\frac{25}{4}$ gives $\frac{1}{4}$ on both sides. The third line is the same numbers written as squares, the way you complete the square: $(2 - \frac{5}{2})^2 = 4 - 10 + \frac{25}{4}$. All true.

Then the square root is taken of both sides. Put numbers in: $2 - \frac{5}{2} = -\frac{1}{2}$ and $3 - \frac{5}{2} = \frac{1}{2}$. So the fourth line says $-\frac{1}{2} = \frac{1}{2}$: false. The rule that broke is “if two squares are equal, the numbers are equal”. It isn’t true: if $x^2 = y^2$, then x = y or x = −y. A number and its opposite have the same square. Squaring is the one-way step from the table, and taking the square root tries to walk back across it.

An equation with an answer that isn’t one

This one looks like ordinary homework, with algebraic fractions. Solve

$$\frac{x}{x - 3} = \frac{3}{x - 3} + 2.$$

Multiply both sides by x − 3: x = 3 + 2(x − 3), so x = 2x − 3, so x = 3. A neat answer. Now check it in the first line: $\frac{3}{0} = \frac{3}{0} + 2$. The equation divides by zero at x = 3; it means nothing there. So 3 is not an answer.

Is there any answer? Take $\frac{3}{x - 3}$ from both sides instead: the left side becomes $\frac{x - 3}{x - 3}$, which is 1 for every allowed x, and the right side is 2. So the equation really says 1 = 2, for every x except 3. It has no solution at all.

What went wrong is the one-way step again. Multiplying by x − 3 is fine for every x but 3; at 3 it is multiplying by 0, which turned a meaningless line into true ones. The method didn’t lie: it said “if there is an answer, it is 3”. There is no answer, so the “if” was false. An answer like this, produced by a step that can’t be undone and failing the first equation, is called an extraneous solution.

The pattern

That is the pattern behind every proof on this page. A false proof is almost never a false rule. It is a true rule used where its condition fails: dividing, with “not by zero” forgotten; square-rooting, with “or its opposite” forgotten; multiplying by x − 3, with “unless it is zero” forgotten. Mathematicians call such a chain a mathematical fallacy, and finding the exact step where it fails, and saying why, is what curricula call critiquing the reasoning.

Here is a twist. Some arguments look just as suspicious, and are perfectly right.

Let x = 0.999…, with nines that never end. Then 10x = 9.999…. Take the first from the second: the endless tails of nines match up and vanish, leaving 9x = 9, so x = 1. Many people refuse to believe this. But it is true: 0.999… = 1. (The 6–8 topic Can You Add Halves Forever and Still Get One? shows why, by the running totals.)

Now use the same trick on S = 1 + 2 + 4 + 8 + …, doubling for ever. Everything after the first 1 is twice the whole: S = 1 + 2S. Take S from both sides: 0 = 1 + S, so S = −1. Adding up positive numbers has given a negative one.

The same kind of steps; one right, one absurd. Where is the broken step this time? Not in the algebra: every step from S = 1 + 2S onward is fine. It is in the first line, “let S be the sum”. Algebra’s rules are rules about numbers, and they can only be used on something that is a number.

The totals of 0.9 + 0.09 + 0.009 + … are 0.9, 0.99, 0.999, …, each ten times closer to 1 than the last. They settle on 1, so there really is a number there, and treating it as one, multiplying it by 10 and taking it away, is safe. The totals of 1 + 2 + 4 + … are 1, 3, 7, 15, …, $2^n - 1$ after n terms: they pass every number and settle nowhere. There is no number S, and the trick answered a question that has no answer.

The third sum, 1 − 1 + 1 − 1 + …, is the strangest. The trick says S = 1 − S, so $S = \frac{1}{2}$. Grouping it as (1 − 1) + (1 − 1) + … gives 0; grouping it as 1 + (−1 + 1) + (−1 + 1) + … gives 1. Its totals go 1, 0, 1, 0 for ever. The Italian priest and mathematician Guido Grandi wrote about it in 1703, and it is still called Grandi’s series. Leibniz and Euler argued for ½. Today a sum whose totals never settle is said to have no sum in the ordinary sense; special methods, such as averaging the running totals (Cesàro’s, 1890), give it ½, but ordinary algebra done on it can be made to give different answers.

Finding the exact step where an argument fails is a skill England’s national curriculum and the United States’ Common Core standards ask of secondary students: to “assess the validity of an argument”, and, “if there is a flaw in an argument”, to “explain what it is”. Outside the classroom, the rule about zero matters most in computers.

• Can You Prove √2 Is Not a Fraction? A proof that works by assuming the opposite and reaching something impossible: the honest use of “absurd”.
• Can a Sentence Say It Is False? When the trouble is not a step but the sentence itself.
• Where Did the Missing Square Go? (Grades 6–8) False proofs made of pictures, where the broken step is “they look the same”.
• Is a Pattern That Works Ten Times Always True? (Grades 6–8) Why checking examples can never replace a proof.
• How Did a Rounding Error Stop a Missile Defense? (Computing, Going Further, Grades 9–12) What happens when a computer’s numbers break the rules of ordinary arithmetic.
• Grade 10: Algebraic fractions The rule x ≠ 3, and canceling factors safely.