Generalize and Specialize

In one sentence

Two moves in opposite directions often work together:

Specialize: if the general case is hard to see, look at the most special one first: the smallest $n$ , the most symmetric position, the simplest numbers. It tells you roughly what the answer is and which way to go with the proof.
Generalize: if the specific numbers are big and messy, replace them with letters. The numbers lose their “personality”, and the structure shows through.

What specializing finds is only a conjecture; the proof still has to deal with the general case.

Predict first

An equilateral triangle $A B C$ is inscribed in a circle, and the point $P$ is on the arc $B C$ that does not contain $A$ . When $P$ is exactly the midpoint of this arc, $P A$ is how many times $P B$ ?

times

Choose an answer first; then you can see the result.

Stop 1 · Scout with special positions (Vol. 5)

Example: how are PA, PB and PC related?

Step 1 / 5

The most symmetric position
Put $P$ at the midpoint of the arc. $P A$ becomes a diameter, and $P B = P C =$ the radius. So $P A = P B + P C$ .

$2 R = R + R$
The most extreme position
Write a conjecture, then look at a general position
Generalize: a special case of Ptolemy’s theorem
It works without Ptolemy too

Going the other way, specializing a general theorem often gives you an old friend. Apply Ptolemy’s theorem to a rectangle: both diagonals are $d$ , and the two pairs of opposite sides are $a$ and $b$ , so $d \cdot d = a \cdot a + b \cdot b$ . That is the Pythagorean theorem.

Stop 2 · Replace numbers with letters (Vol. 2)

Compute $202 3^{2} - 2022 \times 2024$ . Brute force means two multiplications with seven-digit answers.

Example: 2023² − 2022 × 2024

Step 1 / 4

Specialize: try small numbers first
Replace $2023$ by smaller numbers: $2^{2} - 1 \times 3 = 1$ , $3^{2} - 2 \times 4 = 1$ , $4^{2} - 3 \times 5 = 1$ .

$1, 1, 1$
It always seems to be 1
Generalize: use a letter
Back to the problem

Stop 3 · Substitute special values to find factors and coefficients (Vol. 1)

Factor $a^{2} (b - c) + b^{2} (c - a) + c^{2} (a - b)$ . Fully expanded it has six terms and no obvious pattern. Try giving the letters special values.

Example: factor a²(b−c) + b²(c−a) + c²(a−b)

Step 1 / 5

Specialize: set a = b
Replace $a$ by $b$ : the first two terms cancel and the third is $0$ . The expression becomes $0$ , which shows that $(a - b)$ is a factor.

$b^{2} (b - c) + b^{2} (c - b) + c^{2} \cdot 0 = 0$
Rotate the letters
Count the degrees
Specialize again: numbers fix k
The result

Stop 4 · Start counting on a small board (Vol. 7)

How many squares (of every size) are there on an $8 \times 8$ board? Counting them directly will make your eyes swim.

Example: squares on a board

Step 1 / 4

Specialize: the smallest board
A $1 \times 1$ board: $1$ square. A $2 \times 2$ board: $4$ small ones plus $1$ big one, $5$ in all.

$1, 4 + 1 = 5$
A bit bigger
Generalize: why squares?
Back to 8 × 8

Practice

Exercise 1Basic

Compute $999 \times 1001 - 998 \times 1002$ .

Exercise 2Advanced

For all $x$ , $(2 x - 1)^{3} = a_{3} x^{3} + a_{2} x^{2} + a_{1} x + a_{0}$ . Find $a_{0} - a_{1} + a_{2} - a_{3}$ .

Exercise 3Challenge

How many rectangles (squares count as rectangles too) are there on a $6 \times 6$ board?

Mistake Museum · Exhibits from this stop

Before you leave, look at the traps other people fell into here.See every exhibit in the museum →

Exhibit

True in a special case, so you think it is proved

When

P

is the midpoint of arc

B C

P A = P B + P C

, so

P A = P B + P C

for every point

P

on the arc.

What went wrong? Think first, then open

Diagnosis Only one position was checked. Something true in a special position may well be false somewhere else. For example, “the altitude and the median coincide” is true in an isosceles triangle but false in a general one.

Specializing helps you find out what to prove. Once you know, write a proof for

P

in a general position (here: Ptolemy’s theorem, or cut off

P D = P B

P A

“Special values find the answer; they do not prove it”

Exhibit

Too special

To find

k

a^{2} (b - c) + b^{2} (c - a) + c^{2} (a - b) = k (a - b) (b - c) (c - a)

, take

a = b = 1

c = 0

: both sides are

0

, so

k

can be anything.

What went wrong? Think first, then open

Diagnosis These values make both sides $0$ at once, and $0 = k \cdot 0$ holds for every $k$ , so you learn nothing.

Choose values that make both sides nonzero, for example

a = 0, b = 1, c = 2

: the left side is

- 2

and the right side is

2 k

, so

k = - 1

“Be special, but not so special that both sides become 0”

In one sentence

Stop 1 · Scout with special positions (Vol. 5)

Stop 2 · Replace numbers with letters (Vol. 2)

Stop 3 · Substitute special values to find factors and coefficients (Vol. 1)

Stop 4 · Start counting on a small board (Vol. 7)

Practice

Mistake Museum · Exhibits from this stop

Stops on this line