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?
. The bound comes from “drawers”, and the example from “the larger half”.
Example: at most how many numbers from {1, …, 2n}, none dividing another (n = 10)?
Step 1 / 5- 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- 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- 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).
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
At most how many numbers can be chosen from so that none of them divides another?
At most how many numbers can be chosen from so that no two of them are coprime?
Five consecutive positive integers are all composite. What is the smallest possible value of the smallest one?
Real numbers satisfy . Find the minimum of .