Mathematical Induction - Fordham University€¦ · Equating the right-hand sides of equations...

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Mathematical Induction


is a method for proving that:

a property defined for integers n is true

for all values of n that are greater than or equal to some initial integer.


is a method for proving that:

a property defined for integers n is true

for all values of n that are greater than or equal to some initial integer.

ExampleIf we get rid of pennies (1 cent) and replace with new coins of 3 cents value,

With only 3-cent coins and 5-cent coins, can we still give changes of n cents for any positive integer n>7?

n=8: 8 = 3+5 n=9: 9 = 3+3+3 n=10: 10=5+5 n=11: 11=3+3+5 n=12: 12=3+3+3+3 n=13: … How do you convince/prove for any n>7?

Ideas With only 3-cent coins and 5-cent coins, can we still give changes of n cents for any positive integer n>7?

If we can give changes of k cents, how to create solution to (k+1) cents from it?

n=8: 8 = 3+5 n=9: 9 = 3+3+3 n=10: 10=5+5 n=11: 11=3+3+5 n=12: 12=3+3+3+3 n=13: … any pattern?

Ideas: change reaction

if k>=9, then either we use a 5, or we use 3 3-cents.

Ideas: change reaction • We can give change of 9 cents.

• for any k>=9, if we can give change of k cents, then we can give change of k+1 cents.

* So we can give change of 10 cents. * why?

* So we can give change of 11 cents. * why?

* Continue forever…

Logic behind: change reaction • We can give change of 9 cents.

• for any k>=9, if we can give change of k cents, then we can give change of k+1 cents.

* So we can give change of 10 cents. * Universal Modus Ponens

* So we can give change of 11 cents. * Universal Modus Ponens

* Continue forever…

p(9)

for any k>=9, if P(k) then P(k+1)

Mathematical Induction: two steps

Example 1 – Sum of the First n Integers

Use mathematical induction to prove that

Solution:first identify property P(n). Here, P(n) i:

[To see that P(n) is a sentence, its subject being “the sum of the integers from 1 to n” and its verb is “equals.”]

Example 1 – SolutionBasis step: show that the property is true for n = 1, i.e., P(1) is true.

P(1) is obtained by substituting 1 in place of n in P(n). The left-hand side of P(1) is the sum of all the successive integers starting at 1 and ending at 1. This is just 1. Thus P(1) is

cont’d

Example 1 – Solutionthis equation is true because the right-hand side is

which equals the left-hand side.

cont’d

Example 1 – SolutionInductive step: assume that P(k) is true, for a particular but arbitrarily chosen integer k with k ≥ 1. You must then show that P(k + 1) is true. What are P(k) and P(k + 1)? P(k) is obtained by substituting k for every n in P(n). Thus P(k) is

cont’d

Thus P(k + 1) is

or, equivalently,

Example 1 – SolutionIn this case, left-hand side of P(k + 1) is

1 + 2 +· · ·+ (k + 1), which equals (1 + 2 +· · ·+ k) + (k + 1)

But by substitution from the inductive hypothesis,

cont’d

Example 1 – Solutioncont’d

Example 1 – SolutionSo the left-hand side of P(k + 1) is .

Now the right-hand side of P(k + 1) is Thus the two sides of P(k + 1) are equal to each other, and so the equation P(k + 1) is true.

cont’d

a. Evaluate 2 + 4 + 6 +· · ·+ 500.

b. Evaluate 5 + 6 + 7 + 8 +· · ·+ 50.

c. For an integer h ≥ 2, write 1 + 2 + 3 +· · ·+ (h – 1) in closed form.

Applying the Formula for the Sum of the First n Integers

Geometric sequence: each term is obtained from the preceding one by multiplying by a constant factor.

for example: 1, r, r 2, r

3, . . . , r n, . . . .

sum of first n terms of this sequence is given by the formula

for all integers n ≥ 0 and real numbers r not equal to 1.


in expanded form:

and because r 0 = 1 and r

1 = r, the formula for n ≥ 1 can be rewritten as


Prove that , for all integers n ≥ 0 and all real

numbers r except 1.

Solution:In this example property P(n) is again an equation, although in this case it contains a real variable r :

Sum of a Geometric Sequence

Because r can be any real number other than 1, the proof begins by supposing that r is a particular but arbitrarily chosen real number not equal to 1.

Then the proof continues by mathematical induction on n, starting with n = 0.

In the basis step, you must show that P(0) is true; that is, you show the property is true for n = 0.


So you substitute 0 for each n in P(n):

Is P(0) true?

In inductive step, you suppose k is any integer with k ≥ 0 for which P(k) is true; that is, you suppose the property is true for n = k.


So you substitute k for each n in P(n):

Then you show that P(k + 1) is true; that is, you show the property is true for n = k + 1.

So you substitute k + 1 for each n in P(n):


Or, equivalently,

In inductive step, we use another common technique for showing that an equation is true:

We start with the left-hand side, transform itstep-by-step into the right-hand side, using the inductive hypothesis together with algebra and other known facts.


Proof (by mathematical induction): Suppose r is a particular but arbitrarily chosen real number that is not equal to 1, and let the property P(n) be the equation

We must show that P(n) is true for all integers n ≥ 0. We do this by mathematical induction on n.


Show that P(0) is true:

To establish P(0), we must show that

The left-hand side of this equation is r 0 = 1 and the right-

hand side is

also because r 1 = r and r ≠ 1. Hence P(0) is true.


Show that for all integers k ≥ 0, if P(k) is true then P(k + 1) is also true:

[Suppose that P(k) is true for a particular but arbitrarily chosen integer k ≥ 0. That is:] Let k be any integer with k ≥ 0, and suppose that

[We must show that P(k + 1) is true. That is:] We must show that


Or, equivalently, that

[We will show that the left-hand side of P(k + 1) equals the right-hand side.] The left-hand side of P(k + 1) is


which is the right-hand side of P(k + 1) [as was to be shown.] [Since we have proved the basis step and the inductive step, we conclude that the theorem is true.]


Proving an Equality

Two different ways to show that an equation is true:

1. transforming the LHS (left-hand side) and RHS (right-hand side) independently until they are seen to be equal, and

2. transforming one side of the equation until it is the same as the other side of the equation.

3. another approach: show LHS >= RHS, and RHS>=LHS, and then they have to equal to each other

Proving an Equality

Sometimes people use a method that they believe proves equality but that is actually invalid. e.g., to prove basis step for Theorem 5.2.3:

The problem with this method is that starting from a statement and deducing a true conclusion does not prove that the statement is true.

Proving an Equality

What’s going on here?

A true conclusion can also be deduced from a false statement. For instance, the steps below show how to deduce the true conclusion that 1 = 1 from the false statement that 1 = 0:

Proving an Equality

Let

Then

and so

Alternative way to prove

But

Equating the right-hand sides of equations (5.2.1) and (5.2.2) and dividing by r – 1 gives

This derivation of the formula is attractive and is quite convincing.

However, it is not as logically airtight as the proof by mathematical induction.

Deducing Additional Formulas

Proving a Divisibility Property

Use mathematical induction to prove that for all integers n ≥ 0, 22n – 1 is divisible by 3.

Solution:What is P(n)?

I P(n) is the sentence

Example 1 – SolutionBy substitution, the statement for the basis step, P(0), is

The supposition for the inductive step, P(k), is

and the conclusion to be shown, P(k + 1), is

cont’d

Example 1 – SolutionWe know that an integer m is divisible by 3 if, and only if, m = 3r for some integer r.

Now the statement P(0) is true because 22 ● 0 – 1 = 20 – 1 = 1 – 1 = 0, which is divisible by 3 because 0 = 3 ● 0.

To prove the inductive step, you suppose that k is any integer greater than or equal to 0 such that P(k) is true.

This means that 22k – 1 is divisible by 3. You must then prove the truth of P(k + 1). Or, in other words, you must show that 22(k+1) – 1 is divisible by 3.

cont’d

Example 1 – SolutionBut

The aim is to show that this quantity, 22k ● 4 – 1, is divisible by 3. Why should that be so? By the inductive hypothesis,22k – 1 is divisible by 3, and 22k ● 4 – 1 resembles 22k – 1.

cont’d

Example 1 – SolutionObserve what happens, if you subtract 22k – 1 from 22k ● 4 – 1:

Adding 22k – 1 to both sides gives

Both terms of the sum on the right-hand side of this equation are divisible by 3; hence the sum is divisible by 3.

cont’d

Example 1 – SolutionTherefore, the left-hand side of the equation is also divisible by 3, which is what was to be shown.

This discussion is summarized as follows:

Proof (by mathematical induction):

Let the property P(n) be the sentence “22n – 1 is divisible by 3.”

cont’d

Example 1 – SolutionShow that P(0) is true:


But

and 0 is divisible by 3 because 0 = 3 ● 0.

Hence P(0) is true.

cont’d

Example 1 – SolutionShow that for all integers k ≥ 0, if P(k) is true then P(k + 1) is also true:

[Suppose that P(k) is true for a particular but arbitrarily chosen integer k ≥ 0. That is:]

Let k be any integer with k ≥ 0, and suppose that

By definition of divisibility, this means that

cont’d

Example 1 – Solution[We must show that P(k + 1) is true. That is:] We must show that

But

cont’d

Example 1 – Solution

But is an integer because it is a sum of products of integers, and so, by definition of divisibility, is divisible by 3 [as was to be shown].

[Since we have proved the basis step and the inductive step, we conclude that the proposition is true.]

cont’d

Example 2 – Proving an InequalityUse mathematical induction to prove that for all integers n ≥ 3,

Solution:In this example the property P(n) is the inequality

By substitution, the statement for the basis step, P(3), is

Example 2 – SolutionThe supposition for the inductive step, P(k), is

and the conclusion to be shown is

To prove the basis step, observe that the statement P(3) is true because 2 ● 3 + 1 = 7, 23 = 8, and 7 < 8.

cont’d

Example 2 – SolutionTo prove the inductive step, suppose the inductive hypothesis, that P(k) is true for an integer k ≥ 3.

This means that 2k + 1 < 2k is assumed to be true for a particular but arbitrarily chosen integer k ≥ 3.

Then derive the truth of P(k + 1). Or, in other words, show that the inequality is true. But by multiplying out and regrouping,

and by substitution from the inductive hypothesis,

cont’d

Example 2 – SolutionHence

If it can be shown that 2k + 2 is less than 2k+1, then the desired inequality will have been proved.

But since the quantity 2k can be added to or subtracted from an inequality without changing its direction,

And since multiplying or dividing an inequality by 2 does not change its direction,

cont’d

Example 2 – SolutionThis last inequality is clearly true for all k ≥ 2. Hence it is true that .

This discussion is made more flowing (but less intuitive) in the following formal proof:

Proof (by mathematical induction):

Let the property P(n) be the inequality

cont’d

Example 2 – SolutionShow that P(3) is true:


But

Hence P(3) is true.

cont’d

Example 2 – SolutionShow that for all integers k ≥ 3, if P(k) is true then P(k + 1) is also true:

[Suppose that P(k) is true for a particular but arbitrarily chosen integer k ≥ 3. That is:]

Suppose that k is any integer with k ≥ 3 such that

[We must show that P(k + 1) is true. That is:] We must show that

cont’d

Example 2 – SolutionOr, equivalently,

But

[This is what we needed to show.] [Since we have proved the basis step and the inductive step, we conclude that the proposition is true.]

cont’d

A Problem with Trominoes

A Problem with TrominoesA particular type of polyomino, called a tromino, is made up of three attached squares, which can be of two types:

Call a checkerboard that is formed using m squares on a side an m × m (“m by m”) checkerboard.

if one square is removed from a 4 × 4 checkerboard, the remaining squares can be completely covered by L-shaped trominoes.

A Problem with TrominoesFor instance, a covering for one such board is illustrated in the figure below.

It is a beautiful example of an argument by mathematical induction.

A Problem with Trominoes

The main insight leading to a proof of this theorem is the observation that because , when a board is split in half both vertically and horizontally, each half side will have length 2k and so each resulting quadrant will be a checkerboard.

A Problem with TrominoesProof (by mathematical induction):

Let the property P(n) be the sentence

If any square is removed from a 2n ⋅ 2n checkerboard,then the remaining squares can be completely covered by L-shaped trominoes.

Show that P(1) is true:

A 21 ⋅ 21 checkerboard just consists of four squares. If one square is removed, remaining squares form an L, which can be covered by a single L-shaped tromino, as illustrated in the figure to the right. Hence P(1) is true.

A Problem with TrominoesShow that for all integers k ≥ 1, if P(k) is true then P(k + 1) is also true:

[Suppose that P(k) is true for a particular but arbitrarily chosen integer k ≥ 3. That is:] Let k be any integer such that k ≥ 1, and suppose that

If any square is removed from a 2k ⋅ 2k checkerboard, then the remaining squares can be completely covered by L-shaped trominoes.

A Problem with TrominoesP(k) is the inductive hypothesis. [We must show that P(k + 1) is true. That is:] We must show that

If any square is removed from a 2k+1 ⋅ 2k+1 checkerboard, then the remaining squares can be completely covered by L-shaped trominoes.

A Problem with TrominoesConsider a 2k+1 ⋅ 2k+1 checkerboard with one square removed. Divide it into four equal quadrants: Each will consist of a 2k × 2k checkerboard.

In one of the quadrants, one square will have been removed, and so, by inductive hypothesis, all the remaining squares in this quadrant can be completely covered by L-shaped trominoes.

The other three quadrants meet at the center of the checkerboard, and the center of the checkerboard serves as a corner of a square from each of those quadrants.

A Problem with TrominoesAn L-shaped tromino can, therefore, be placed on those three central squares. This situation is illustrated in the figure to the right.

By inductive hypothesis, the remaining squares in each of the three quadrants can be completely covered by L-shaped trominoes.

Thus every square in the 2k+1 ⋅ 2k+1 checkerboard except the one that was removed can be completely covered by L-shaped trominoes [as was to be shown].

Copyright © Cengage Learning. All rights reserved.

Strong Mathematical Inductionand the Well-Ordering Principle for

the Integers

Strong Mathematical Induction

similar to ordinary mathematical induction • a technique for proving of a sequence of statements

about integers. • also consists of a basis step and an inductive step.

Difference • basis step may contain proofs for several initial values,

• inductive step: prove: for any integer k>=a, if P(a), P(a+1) … P(k), then P(k + 1)


also known as second principle of induction, second principle of finite induction, and principle of complete induction.


Any statement that can be proved with ordinary math. induction can be proved with strong math. induction.

given any integer k ≥ b, if the truth of P(k) alone implies the truth of P(k + 1), then certainly the truth of P(a), P(a + 1), . . . , and P(k) implies the truth of P(k + 1).


Any statement that can be proved with strong math. induction can be proved with ordinary math. induction.

Strong Math. Induction Ord. Math Induction Let Q(n)=

• P(a), P(a+1), … P(b) * Q(b)

• for any k>=b, if * for any k>=b, if Q(k) then P(a), P(a+1), … P(k) are Q(k+1) true, then P(k+1) is true

P (a) ^ P (a+ 1) ^ ...P (n� 1) ^ P (n)

Applying Strong MathematicalInduction

Example 1 – Divisibility by a PrimeProve: Any integer greater than 1 is divisible by a prime number.

Solution:

I inductive step? If a given integer greater than 1 is not itself prime, then it is a product of two smaller positive integers, each of which is greater than 1.

Since you are assuming that each of these smaller integers is divisible by a prime number, by transitivity of divisibility, those prime numbers also divide the integer you started with.

Example 1 – SolutionProof (by strong mathematical induction):

Let the property P(n) be the sentence n is divisible by a prime number.

Show that P(2) is true: To establish P(2), we must show that 2 is divisible by a prime number.

this is true because 2 is divisible by 2 and 2 is a prime number.

cont’d

Example 1 – Inductive stepShow that for all integers k ≥ 2, if P(i ) is true for all integers i from 2 through k, then P(k + 1) is also true:

Let k be any integer with k ≥ 2 and suppose that i is divisible by a prime number for all integers i from 2 through k.

We must show that k + 1 is divisible by a prime number.

cont’d

Example 1 – SolutionCase 1 (k + 1 is prime): k + 1 is divisible by a prime number, namely itself.

Case 2 (k + 1 is not prime): k + 1 = ab where a and b are integers with 1 < a < k + 1 and 1 < b < k + 1.

as 2 ≤ a ≤ k, and so by inductive hypothesis, a is divisible by a prime number p.

In addition because k + 1 = ab, we have that k + 1 is divisible by a.

cont’d

Example 1 – Solution a| (k + 1) and p| a , so by transitivity of divisibility, p| (k + 1)

Therefore, regardless of whether k + 1 is prime or not, it is divisible by a prime number [as was to be shown].

[Since we have proved both the basis and the inductive step of the strong mathematical induction, we conclude that the given statement is true.]

cont’d

Well-Ordering Principle for the Integers


• Well-ordering principle for integers • ordinary principles of mathematical induction, • strong principles of mathematical induction

but it can be shown that all three principles are equivalent, i.e., if any one of the three is true, then so are both of the others.


Prove that any integer greater than 1 is divisible by a prime number. Proof: Suppose not, i.e., there exist some integers greater than 1 that is not divisible by a prime. Let C be the set of all integers greater than one that is not divisible by any primes, so C is not empty. By well-ordering principle, there exists a least element in C, call it n. n cannot be prime (as if it’s prime, then n|n, and n is not in C) so n = ab, where 1<a,b<n, therefore a, b are not in C so a and b are divisible by some prime numbers p, q, by transitivity of divisibility, p|n, then n is not in C.


Prove that any integer greater than 1 is divisible by a prime number. Proof: Suppose not, i.e., there exist some integers greater than 1 that is not divisible by a prime. Let C be the set of all integers greater than one that is not divisible by any primes, so C is not empty. By well-ordering principle, there exists a least element in C, call it n. [we will derive a contradiction]

Finding Least Elements

Does the set has a least element? If so, what is it? If not, explain why the well-ordering principle is not violated.

a. the set of all positive real numbers.


In each case, if the set has a least element, state what it is. If not, explain why the well-ordering principle is not violated.

b. the set of all nonnegative integers n such that n2 < n.

Example 4 – Solution

b. There is no least nonnegative integer n such that n2 < n because there is no nonnegative integer that satisfies this inequality.

The well-ordering principle is not violated because the well-ordering principle refers only to sets that contain at least one element.

cont’d


In each case, if the set has a least element, state what it is. If not, explain why the well-ordering principle is not violated.

c. The set of all nonnegative integers of the form 46 – 7k, where k is an integer.

{ 46-7k | k is an integer}

46 is an element…


by subtracting six 7’s from 46 leaves 4 left over This is the least nonnegative integer obtained by repeated subtraction of 7’s from 46.

• 6 is the quotient and • 4 is the remainder for the division of 46 by 7.

More generally, in the division of any integer n by any positive integer d, the remainder r is the least nonnegative integer of the form n – dk.


Proof:

Let S be the set of all nonnegative integers of the form

where k is an integer.

The Well-Ordering Principle for the Integers

This set has at least one element. [For if n is nonnegative, then

and so n – 0 ● d is in S. And if n is negative, then

and so n – nd is in S.] By well-ordering principle, S contains a least element r. Then, for some specific integer k = q,

[because every integer in S can be written in this form].

The Well-Ordering Principle for the Integers

Adding dq to both sides gives

Furthermore, r < d. [For suppose r ≥ d.

Then

and so n – d(q + 1) would be a nonnegative integer in S that would be smaller than r. But r is the smallest integer in S. This contradiction shows that the supposition r ≥ d must be false.]


The preceding arguments prove that there exist integers r and q for which

[This is what was to be shown.]

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