Lectoescritura

The previous example needed the truth of P(

u)

and P(

u

-

I ) .

The next example uses the truth of P(k) for two arbitrary values.

Example 2.3.9 A partition of a set is a decomposition of the set into disjoint subsets.

A triangulation of a polygon is a partition of the polygon into triangles, all of whose vertices are vertices of the original polygon. A given polygon can have many different triangulations. Here are two different triangulations of a 9-gon.

Given a triangulation, call two vertices adjacent if they are joined by the edge of a triangle. Suppose we decide to color the vertices of a triangulated polygon. How many colors do we need to use in order to guarantee that no two adjacent vertices have the same color? Certainly we need at least three colors, since that is required for just a single triangle. The surprising fact is that

Three colors always suffice for any triangulation of a polygon!

Proof" We shall induct on

n,

the number of vertices of the polygon. The statement P(

n

) that we wish to prove for all integers

n

2: 3 is

For any triangulation of an n-gon, it is possible to

^3-color

the vertices (i.e., color the vertices using three colors) so that no two adjacent ver­

tices are the same color.

The base case P(3) is obviously true. The inductive hypothesis is that P (3 ) , P(4) , . . . ,P (

n

)

are all true. We will show that this implies P(

n

+

1 ) .

Given a triangulated

(n

+

1 ) ­

gon, pick any edge and consider the triangle

T

with vertices

x, y, z

that contains this edge. This triangle cuts the

(n

+

1

) -gon into two smaller triangulated polygons L and

2.3 M ETHODS OF A R G U M ENT 49 R (abbreviating left and right, respectively). It may tum out that one of L or R doesn't exist (this will happen if n =

4),

but that doesn't matter for what follows. Color the vertices of L red, white, and blue in such a way that none of its adjacent vertices are the same color. We know that this can be done by the inductive hypothesis !

Note that we have also colored two of the vertices

(x

and

y)

of

T ,

and one of these vertices is also a vertex of R. Without loss of generality, let us assume that x is blue and

y

is white.

By the inductive hypothesis, it is possible to 3-color R using red, white and blue so that no two of its vertices are the same color. But we have to be careful, since R shares two vertices (y ^and

z)

^with

T,

one of which is already colored white. Consequently, vertex

z

must be red. If we are lucky, the coloring of R will coincide with the coloring of

T.

But what if it doesn 't? No problem. Just rename the colors ! In other words, interchange the roles of red, blue, and white in our coloring of R. For example, if the original coloring of R made

y

red and

z

blue, just recolor all of R 's red vertices to be white and all blue vertices red.

We 're done; we 've successfully 3-colored an arbitrarily triangulated (n + I ) -gon so that no two adjacent vertices are the same color. So P{n +

1 )

is true. _

T

·

· ·

·

\ i ,,'/

\,:,,1

Notice that we really needed the stronger inductive hypothesis in this proof. Per­

haps we could have arranged things as in Example 2.3.5 on page 45 so that we decom­

posed the (n +

1

) -gon into a triangle and an n-gon. But it is not immediately obvious that among the triangles of the triangulation, there is a "boundary" triangle that we can pick. It is just easier to pick an arbitrary edge, and assume the truth of P {k) for

all k

� n.

Behind the idea of strong induction is the notion that one should stay flexible in defining hypotheses and conclusions. In the following example, we need an unusually strong inductive hypothesis in order to make progress.

Example 2.3.10 Prove that

Solution : We begin innocently by letting P{n) denote the above assertion.

P( I )

is true, since I /² � 1 /

V3.

Now we try to prove

P(n

+

I ) ,

given the inductive hypothesis that P{n) is true. We would like to show that

{ (

₂

¹ ) (

_{4 · · ·} ³

) (

_2;;-²ⁿ

^{- I} ) } (

^{2n +}

^I )

_� ¹

Extracting the quantity in the large brackets, and using the inductive hypothesis, it will suffice to prove

1 ( ^{2n + l} ) ¹

y3n 2n + 2

:S

Unfortunately, this inequality is false ! For example, if you plug in

n

⁼

I ,

you get

1 ( ³ ) ¹

v'3

4

:S J6 ' which would imply

V2

:s

4/3,

which is clearly absurd.

What happened? We employed wishful thinking and got burned. It happens from time to time. The inequality that we wish to prove, while true, is very weak (i.e., asserts very little), especially for small

n.

The starting hypothesis of

P( I )

is too weak to lead to

P(2) ,

and we are doomed.

The solution: strengthen the hypothesis from the start. Let us replace the

3n

with

3n + I .

Denote the statement

(�) (�)

^{. .}

^{. I} )

^:s

by

Q(n) .

Certainly,

Q( I )

is true; in fact, it is an

equality ( 1 /2

=

1 / v4),

which is about as sharp as an inequality can be ! So let us try to prove

Q(n + I )

using

Q(n)

as the inductive hypothesis. As before, we try the obvious algebra, and hope that we can prove the inequality

1 + 1 2n + 2 ( ^{2n + l} )

:s

¹ + 4 .

Squaring and cross-multiplying reduce this to the alleged inequality

(3n + 4) (4n2 + 4n + I )

:s

4(n2 + 2n + 1 ) (3n + I ) ,

which reduces (after some tedious multiplying) to

1 9n

:s

20n,

and that is certainly true.

So we are done. _

Problems and Exercises

2.3.1 1 Let

a,b,c

be integers satisfying

a2 +b2

⁼

c2•

Give two different proofs that

abc

must be even, (a) by considering various parity cases;

(b) using argument by contradiction.

2.3.12 Make sure you understand Example 2.3.2 per­

fectly by doing the following exercises.

(a) Prove that J3 is irrational.

(b) Prove that v'6 is irrational.

(c) If you attempt to prove J49 is irrational by

us-ing the same argument as before, where does the argument break down?

2.3.13 Prove that there is no smallest positive real number.

2.3.14 Prove that log 10 2 is irrational . 2.3.15 Prove that v'2

+

^J3 is irrational.

2.3.16 Can the complex numbers be ordered? In other words, is it possible to define a notion of "inequality"

so that any two complex numbers

a+bi

^and

c+di

^can

be compared and it can be decided that one is "bigger"

or one is "smaller" or they are both equal? (Using the norm function

Ix + iyl = X2 +

y2 is "cheating," for this converts each complex number into a real num­

ber and hence eludes the question of whether any two complex numbers can be compared.)

2.3.17 Prove that if

a

is rational and

b

is irrational, then

a + b

is irrational.

2.3.18 True or false and why: If

a

^and

b

are irrational, then

ab

is irrational.

2.3.19 Prove the statements made in the discussion of congruence notation on page 44.

2.3.20 Prove the following generalization of Exam­

ple 2.3.4 on page 44:

Let m be a positive integer and let S de­ note the set of positive integers less than

In document Eficacia de pictogramas como estrategia pedagógica para mejorar la lectoescritura en estudiantes de primer grado de Primaria Lima-2020 (página 32-35)