Method line M18 · When and why do we use it?

Construction

In the Chinese book: 构造法

Build what you are looking for: an example, a function, a figure.

📖 Chapter 18 of 《解题方法与策略》

One idea, one pictureChoose as many numbers from 1 to 20 as you can so that none of them divides another. How many can you pick? Then tap “Group into drawers” to see why you cannot do better.

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In one sentence

The answer to many problems is a “thing”: an example, a function, a figure. Construction means building it. The most common pattern is a max/min problem:

With only the bound, the answer might be smaller; with only the example, the answer might be larger. Another kind of construction is “recognizing” an expression: if it looks like a discriminant, build a quadratic function; if it looks like the law of cosines, build a triangle.

Predict first

At most how many numbers can be chosen from so that none of them is a multiple of another?

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

Example: at most how many numbers from {1, …, 2n}, none dividing another (n = 10)?

Step 1 / 5
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  1. Pick a few at random

    Greedily picking from small to large is a bad idea: once is chosen, are all ruled out. The detector marks every pair where one divides the other.

Stop 1 · Constructing a function (Vol. 2)

has the shape “”, like a discriminant. So build a quadratic function whose discriminant it is.

One problem, several solutions

Prove the Cauchy inequality .

Construct a quadratic function

Assume (otherwise it is obvious). Let . Then for every real and the parabola opens upward, so the discriminant satisfies . Equality holds if and only if some gives for all .

CostYou first have to “see” the discriminant; after that, two lines.

Lagrange's identity

.

CostJust expand to check, but you have to remember the identity.

Stop 2 · Constructing a figure (Vol. 3)

is exactly what the law of cosines looks like with an angle of : .

Example: a, b, c > 0. Prove that √(a² − ab + b²) + √(b² − bc + c²) ≥ √(a² + ac + c²)

Step 1 / 4
  1. Recognize the law of cosines

    ; .

Stop 3 · Existence in sequences (Vol. 5)

“Are there arbitrarily long runs of consecutive composite numbers?” Yes, and you can write them down.

Example: for every positive integer k there are k consecutive composite numbers

Step 1 / 3
  1. Give every term a divisor

    We want the -th number to be divisible by : start from a number divisible by all of .

Stop 4 · Partitions of a set (Vol. 1)

The same pattern with a different condition: the chosen numbers must be pairwise not coprime. The drawers become pairs of neighbors (consecutive integers are coprime).

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Choose as many numbers from to as you can so that no two are coprime. First switch on “Group into drawers”, then build your example.

Upper bound: the drawers each allow at most one choice, so at most numbers. Construction: all the even numbers , any two of which share the factor . The maximum is .

Practice

Exercise 1Basic

At most how many numbers can be chosen from so that none of them divides another?

Exercise 2Basic

At most how many numbers can be chosen from so that no two of them are coprime?

Exercise 3Advanced

Five consecutive positive integers are all composite. What is the smallest possible value of the smallest one?

Exercise 4Challenge

Real numbers satisfy . Find the minimum of .

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

A constructed example that was never checked

The wrong step: Choose numbers from , none dividing another: take all the odd numbers , since odd numbers do not divide each other.
What went wrong? Think first, then open

Diagnosis divides everything, divides … Odd numbers can certainly divide each other. A constructed example must be checked against every condition; “looks right” does not count.

The right step: Take : the ratio of any two of them is greater than and less than , so neither can divide the other.

“Check every condition of your example”

Exhibit

The factorial construction is off by one

The wrong step: To get consecutive composite numbers, take .
What went wrong? Think first, then open

Diagnosis divides only for . Since , there is no guarantee that is composite.

The right step: Take : and , so all ten numbers are composite.

“Every term of a construction needs a reason”

Stops on this line

The same idea, used once in each of several places.

  1. Solving Problems by Constructing a FunctionVol. 2 Functions and Functional Equations, ch. 5
  2. Partitions of a SetVol. 1 Sets, ch. 4
  3. Trigonometric Substitution and ApplicationsVol. 3 Trigonometric Functions, ch. 6
  4. Existence Problems for SequencesVol. 5 Sequences and Mathematical Induction, ch. 15
  5. Proving Inequalities with CauchyVol. 6 The Mean Inequalities and the Cauchy Inequality, §4.1