An argument that looks careful
Here is an argument that proves 2 = 1. Line 1: let a = b. Line 2: multiply both sides by a, so . Line 3: subtract from both sides, so .
Each move is stated, and each one is allowed: multiplying both sides of an equation by the same number, or subtracting the same number from both sides, keeps a true equation true.
An absurd ending
Line 4: factor both sides, so (a + b)(a − b) = b(a − b). Line 5: divide both sides by a − b, so a + b = b. Line 6: since a = b, this says 2b = b, and dividing by b gives 2 = 1.
The conclusion is false, so the argument cannot be sound. Allowed moves carry a true line to a true line, so if every move here were allowed, a true first line could never lead to 2 = 1. At least one move is not allowed, and the task is to say which one.
Put numbers into every line
Pick numbers that satisfy line 1, say a = b = 3, and work out both sides of every line. Line 1 reads 3 = 3. Line 2 reads 9 = 9. Line 3 reads 9 − 9 = 9 − 9, that is 0 = 0.
Line 4 reads (3 + 3)(3 − 3) = 3(3 − 3), that is 6 × 0 = 3 × 0, and both sides are 0. Still true.
Line 5 reads 3 + 3 = 3, that is 6 = 3. False. Line 5 is the first false line, so the move that produced it is the flaw. That move divided both sides by a − b, and a − b = 3 − 3 = 0.
Dividing by 0 is not allowed. 6 × 0 = 3 × 0 is true only because anything times 0 is 0; it says nothing about 6 and 3, just as 1 × 0 = 2 × 0 says nothing about 1 and 2. Undoing the multiplication by 0 would need a division by 0, and no number does that job.
The last move, dividing 2b = b by b, is allowed here, because b = 3 is not 0. It does not matter: line 5 was already false.
Both sides of each line with a = b = 3. The two sides agree in lines 1 to 4 and first disagree in line 5, which came from dividing by a − b = 0.
The moves to check first
A few moves account for most flaws. Check them first when a conclusion is absurd.
Dividing by something that is zero. A letter, or an expression such as a − b, can be zero even when it does not look like it. Before dividing, ask whether it could be 0.
Taking a square root and losing a sign. From it does not follow that x = 3: x can also be −3, since . In the same way is true, and taking square roots of both sides to get −3 = 3 is not. The square root of is |x|, not x.
Using the conclusion as an assumption. To prove that , an argument that starts "since , we have , which is true" has assumed what it set out to show. It has only checked that the claim leads to something true, and a false claim can do that too: from 2 = 1, multiplying both sides by 0 gives the true line 0 = 0.
The gold curve meets the line y = 9 twice, at x = −3 and at x = 3. An argument that goes from to x = 3 has lost the crossing on the left.
Reading an argument
When an argument ends somewhere false, do not hunt in the last line. Choose values that make the first line true and test the lines in order. The first false line points to the move that broke the chain; read that move and ask what it assumed.
A true conclusion is no protection either. An argument can reach a true statement through a broken step, so a proof is checked move by move, not by whether its last line happens to be right.
The usual mistakes
Blaming the line where the absurdity appears. 2 = 1 shows up in line 6, but line 6 follows correctly from line 5; the first false line is line 5.
Blaming the multiplication. Multiplying both sides by a, or even by 0, keeps a true equation true. The flaw is a division by 0, which is the reverse move.
Deciding that every step must be wrong. One bad move is enough. Here every move but one is correct.
Taking the positive root only. has two solutions, 3 and −3.
A café and a game
The first application below is the same algebra with a café’s revenue R and costs C, both $8000, and it finds the first false line by putting the figures in. The second is an argument by induction whose step is correct and whose base case is missing.
Worked example: A Café That Broke Even, and a Nephew's Algebra That Ends in 2 = 1
Question A café broke even last month: its revenue R equaled its costs C, and both were $8000. The owner's nephew writes: line 1, R = C; line 2, multiply both sides by R: R2 = RC; line 3, subtract C2 from both sides: R2 − C2 = RC − C2; line 4, factor both sides: (R + C)(R − C) = C(R − C); line 5, divide both sides by R − C: R + C = C; line 6, replace R by C: 2C = C, so 2 = 1. (a) Put R = C = 8000 into every line. Which is the first false line, and why is the move that produced it not allowed? (b) The nephew then says: 'Line 2 on its own proves a month broke even, because dividing R2 = RC by R gives R = C.' Give the revenue of a month with costs of $8000 for which R2 = RC holds but the café did not break even.
1.Test the first lines with R = C = 8000. Line 1 reads 8000 = 8000. Line 2 reads 64000000 = 64000000. In line 3 both sides are 64000000 − 64000000 = 0.
With R = C = 8000: line 1 reads 8000 = 8000, line 2 reads 64000000 = 64000000, and in line 3 both sides are 64000000 − 64000000 = 0. 2.Line 4: the left side is (8000 + 8000)(8000 − 8000) = 16000 × 0 = 0, and the right side is 8000 × 0 = 0. Lines 1 to 4 are all true.
Line 4: the left side is (8000 + 8000)(8000 − 8000) = 16000 × 0 = 0 and the right side is 8000 × 0 = 0. Lines 1 to 4 are all true. 3.(a) Line 5 reads 8000 + 8000 = 8000, that is 16000 = 8000, so line 5 is the first false line. It came from dividing both sides by R − C, and R − C = 8000 − 8000 = 0. Dividing by 0 is not allowed: 16000 × 0 = 8000 × 0 is true only because both sides are 0, just as 1 × 0 = 2 × 0 is, and it says nothing about 16000 and 8000.
(a) Line 5 reads 16000 = 8000: the first false line. It divided both sides by R − C, which is 0, and dividing by 0 is not allowed: 16000 × 0 = 8000 × 0 only because both sides are 0. 4.The second argument divides R2 = RC by R, which is allowed only when R ≠ 0. So try R = 0: then R2 = 0 and RC = 0 × 8000 = 0, and the equation holds.
Dividing R2 = RC by R is allowed only when R ≠ 0. Try R = 0: then R2 = 0 and RC = 0 × 8000 = 0, and the equation holds. 5.(b) A month with revenue $0 and costs of $8000, such as a month the café was closed for repairs, has R2 = RC but made a loss of $8000, so it did not break even. The correct reading of line 2 is R(R − C) = 0, so R = 0 or R = C.
(b) Revenue of $0 with costs of $8000 satisfies R2 = RC, but the café made a loss of $8000: R(R − C) = 0 means R = 0 or R = C.
Answer: (a) Line 5, which reads 16000 = 8000; it divides both sides by R − C, which is 0, and dividing by 0 is not allowed; (b) revenue of $0: then R2 = RC = 0, but R ≠ C
Common mistakes
- Blaming line 6, where the 2 = 1 first appears. Line 6 follows correctly from line 5; the first false line is line 5, and the move that produced it is the division by R − C = 0.
- Blaming line 2 for multiplying by R. Multiplying both sides by any number, zero included, keeps a true equation true; it is dividing by zero that turns a true line into a false one.
Worked example: A Game Designer's Points for Each Level, and an Argument by Induction That They Split Among Three Players
Question A game designer awards 4n + 1 points for completing level n, for n = 1, 2, 3, …, and wants every award to split equally among the 3 players of a team. She argues: 'Suppose the award at level k is a multiple of 3. The award at level k + 1 is 4k+1 + 1 = 4(4k + 1) − 3, a multiple of 3 minus 3, so it is also a multiple of 3. By induction, every award splits equally.' (a) Which part of a proof by induction is missing from her argument, and what remainder does the award at each level leave on division by 3? (b) She changes the award to 4n + c points, where c is a positive whole number. What is the smallest c for which the same kind of argument, with every part in place, proves that every award splits equally among 3 players?
1.Check the step. If 4k + 1 is a multiple of 3, then 4k+1 + 1 = 4 × 4k + 1 = 4(4k + 1) − 3 is a multiple of 3 minus 3, which is again a multiple of 3. The step is correct.
The step is correct: 4k+1 + 1 = 4(4k + 1) − 3. 2.Check the base case, which she never stated. At level 1 the award is 4 + 1 = 5 points, and 5 is not a multiple of 3. The step only passes the claim on from one level to the next, so with nothing true at level 1 the argument proves nothing.
The base case fails: level 1 gives 5 points, and 5 is not a multiple of 3. 3.(a) The base case is missing, and it is false. 4 leaves remainder 1 on division by 3, so every power 4n does too, and 4n + 1 leaves remainder 1 + 1 = 2 at every level. Check: 5 = 3 + 2, 17 = 15 + 2, 65 = 63 + 2 and 257 = 255 + 2.
(a) The base case is missing and false: every award leaves remainder 2 on division by 3. 4.For 4n + c, the step holds for every whole number c: 4k+1 + c = 4(4k + c) − 3c, a multiple of 3 minus a multiple of 3.
For 4n + c the step holds whatever c is, so only the base case 4 + c decides. 5.(b) So only the base case decides: 4 + c must be a multiple of 3. c = 1 gives 5, which is not, and c = 2 gives 6, which is. The smallest c is 2. Check: 4 + 2 = 6, 16 + 2 = 18 and 64 + 2 = 66 are all multiples of 3.
(b) 4 + c is a multiple of 3 first at c = 2.
Answer: (a) The base case is missing: level 1 gives 5 points, which is not a multiple of 3; every award leaves remainder 2 on division by 3; (b) c = 2
Common mistakes
- Finding the step correct and accepting the proof. A correct step shows only that the claim passes from each level to the next; without a true base case there is nothing to pass on, and here the claim is false at every level.
- Looking for the error in the algebra 4(4k + 1) − 3. That line is correct, since 4 × 4k + 4 − 3 = 4k+1 + 1; the flaw is the missing base case, not the step.