5: Induction - School of Computer Science, University of ...graham/cs1021/slides/induction.pdf · The Principle of Induction Induction 5-1. CS1021 The natural numbers (continued)

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5: Induction

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The natural numbers

N = (N,+,×, 0, 1)

where the elements of N are

0,1,2,3,4,5, . . .

(the non-negative integers).

In many ways this structure is similar to the integers Z, but has an important

extra property.

The Principle of Induction

Induction 5-1

CS1021 The natural numbers (continued)'

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One form of this principle is as follows

The induction principle for natural numbers

Suppose that X is a set of natural numbers such that both

(Base) 0 ∈ X

(Step) ∀n ∈ N, n ∈ X ⇒ n + 1 ∈ X

hold. Then ∀n ∈ N, n ∈ X . In other words X = N

Induction 5-2


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1

2

3

4

5

n

n+1

Induction 5-3


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An alternative formulation is the following

Suppose that P(n) is a property defined for all natural numbers n

such that both

(Base) P(0) is true

(Step) ∀n ∈ N, if P(n) is true, then so is P(n + 1)

hold. Then ∀n ∈ N, P(n) is true.

Induction 5-4


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There are many places where we need to verify a fact of the form

For each natural number n · · ·n · · ·

where · · ·n · · · is some assertion about arbitrary numbers n.

Induction 5-5

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Example – recursive function

Suppose the function

f : N > Z

satisfies

f (0) = 0

f (1) = 1

f (r + 2) = 2 f (r + 1)− f (r)+ 2

∀r ∈ N.

f (0) = 0

f (1) = 1

f (2) = 2−0 + 2 = 4

Induction 5-6

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f (3) = 8−1 + 2 = 9

f (4) = 18−4 + 2 = 16

...

In general, ∀r ∈ N,

f (r) = r2.

How do we prove this?

Induction 5-7

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Example – Divisibility

Consider the expression

f (n) = 32n + 4n+1

n f (n)

0 30 + 41 = 5

1 32 + 42 = 25

2 34 + 43 = 145

3 36 + 44 = 985

4 38 + 45 = 7585

Claim that f (n) is always divisible by 5.

How do we prove it?

Induction 5-8

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Example – sum of a series

Let Sum(n) be the sum of the first n + 1 natural numbers, i.e.

Sum(n) = 0 + 1 + 2 + 3 + · · ·+ n.

Can we find a compact formula for Sum(n)?

Notation:-

s

∑i=r

ai = ar + ar+1 + · · ·+ as

in particular

Sum(n) =n

∑i=0

i.

Induction 5-9

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Three summation problems

Find compact formulas for the following.

(i)n∑

i=0i i.e. 0 + 1 + 2 + 3 · · ·+ n

(ii)n∑

i=0i2 i.e. 0 + 1 + 4 + 9 · · ·+ n2

(iii)n∑

i=0i(i + 1) i.e. 0 + 2 + · · ·+ n(n + 1)

We will show that

Induction 5-10

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n∑

i=0i = n(n+1)

2

n∑

i=0i2 = n(n+1)(2n+1)

6

n∑

i=0i(i + 1) = n(n+1)(n+2)

3

Induction 5-11

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Special ‘trick’ method

Let

Sum(n) =n

∑i=0

i

so that

Induction 5-12

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Let

Sum(n) =n

∑i=0

i

so that

Sum(n) = 0+ 1+ · · · +(n−1) +n

Induction 5-12

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Let

Sum(n) =n

∑i=0

i

so that

Sum(n) = 0+ 1+ · · · +(n−1) +n

Sum(n) = n+ (n−1)+ · · · +1 +0

Induction 5-12

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Let

Sum(n) =n

∑i=0

i

so that

Sum(n) = 0+ 1+ · · · +(n−1) +n

Sum(n) = n+ (n−1)+ · · · +1 +0

Adding gives

Induction 5-12

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Let

Sum(n) =n

∑i=0

i

so that

Sum(n) = 0+ 1+ · · · +(n−1) +n

Sum(n) = n+ (n−1)+ · · · +1 +0

Adding gives

2Sum(n) = n+ n+ · · · +n +n

Induction 5-12

CS1021 Special ‘trick’ method (continued)'

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2Sum(n) = n + n + n + · · ·+ n + n = n(n + 1)

there are n + 1 terms each of value n. So

Induction 5-13


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2Sum(n) = n + n + n + · · ·+ n + n = n(n + 1)


Sum(n) =n(n + 1)

2.

Induction 5-13


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2Sum(n) = n + n + n + · · ·+ n + n = n(n + 1)


Sum(n) =n(n + 1)

2.

This trick won’t work in general. Need a more general method.

Induction 5-13

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A more general method

∀n ∈ N let

L(n) =n

∑i=0

i , R(n) =n(n + 1)

2

so we want to show the following.

∀n ∈ N, L(n) = R(n)

We can verify directly that

Induction 5-14

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L(0) = 0 = R(0)

L(1) = 1 = R(1)

L(2) = 3 = R(2)

L(3) = 6 = R(3)

L(4) = 10 = R(4)...

...

but we need a proof for all n ∈ N.

Induction 5-15

CS1021 A more general method (continued)'

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We can verify directly that:

∀n ∈ N, if

L(n) = R(n)

then also

L(n + 1) = R(n + 1).

i.e. the Induction Step

(Step) L(n) = R(n) ⇒ L(n + 1) = R(n + 1)

and then

Induction 5-16


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(0) L(0) = R(0) by direct computation

(1) L(1) = R(1) from (0), (Step)

(2) L(2) = R(2) from (1), (Step)

(3) L(3) = R(3) from (2), (Step)...

and hence

∀n ∈ N, L(n) = R(n).

Induction 5-17


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L(n + 1)

=n+1∑

i=0i Definition of L

Induction 5-18


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L(n + 1)

=n+1∑


= (n∑

i=0i)+(n + 1) Isolate the last term

Induction 5-18


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L(n + 1)

=n+1∑


= (n∑


= L(n)+(n + 1) Definition of L

Induction 5-18


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L(n + 1)

=n+1∑


= (n∑



= R(n)+(n + 1) Induction Hypothesis

Induction 5-18


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L(n + 1)

=n+1∑


= (n∑




= n(n+1)2 +(n + 1) Definition of R

Induction 5-18


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L(n + 1)

=n+1∑


= (n∑





∗= (n+1)(n+2)

2 Simple arithmetic

Induction 5-18


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L(n + 1)

=n+1∑


= (n∑





∗= (n+1)(n+2)

2 Simple arithmetic

= R(n + 1) Definition of R

Induction 5-18


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L(n + 1)

=n+1∑


= (n∑





∗= (n+1)(n+2)

2 Simple arithmetic


The equality∗= is the only non-routine step.

Induction 5-18


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L(n + 1)

=n+1∑


= (n∑





∗= (n+1)(n+2)

2 Simple arithmetic



This means that L(n) = R(n) ∀n ∈ N

Induction 5-18

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A second example

∀n ∈ N let

L(n) =n

∑i=0

i2 , R(n) =n(n + 1)(2n + 1)

6.

We show the following.

∀n ∈ N, L(n) = R(n)

Induction 5-19

CS1021 A second example (continued)'

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Note first that

L(0) = 0 = R(0).

We now show the Induction Step:

∀n ∈ NL(n) = R(n) ⇒ L(n + 1) = R(n + 1).

Induction 5-20


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L(n + 1)

=n+1∑

i=0i2 Definition of L

Induction 5-21


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L(n + 1)

=n+1∑


= (n∑

i=0i2)+(n + 1)2 Isolate last term

Induction 5-21


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L(n + 1)

=n+1∑


= (n∑


= L(n)+(n + 1)2 Definition of L

Induction 5-21


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L(n + 1)

=n+1∑


= (n∑



= R(n)+(n + 1)2 Induction Hypothesis

Induction 5-21


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L(n + 1)

=n+1∑


= (n∑




= n(n+1)(2n+1)6 +(n + 1)2 Definition of R

Induction 5-21


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L(n + 1)

=n+1∑


= (n∑





∗= (n+1)(n+2)(2n+3)

6 Arithmetic

Induction 5-21


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L(n + 1)

=n+1∑


= (n∑





∗= (n+1)(n+2)(2n+3)

6 Arithmetic


Induction 5-21


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L(n + 1)

=n+1∑


= (n∑





∗= (n+1)(n+2)(2n+3)

6 Arithmetic



Induction 5-21

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A general pattern

∀n ∈ N let

L(n) =n

∑i=0

ai , R(n) = explicit formula.

We wish to show that

∀n ∈ N, L(n) = R(n)

Induction 5-22

CS1021 A general pattern (continued)'

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This follows from

Base : L(0) = R(0)

and

Induction Step: ∀n ∈ N,

L(n) = R(n) ⇒ L(n + 1) = R(n + 1).

Induction 5-23

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Proof of Induction Step

L(n + 1)

=n+1∑

i=0ai Definition of L

= (n∑

i=0ai)+ an+1 Isolate the last term

= L(n)+ an+1 Definition of L

= R(n)+ an+1 Induction Hypothesis∗= R(n + 1) Arithmetic

The equality∗= is the crucial step.

Induction 5-24

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Divisibility problem

Show that if

f (n) = 32n + 4n+1

then f (n) is divisible by 5 ∀n ∈ NBase: f (0) = 30 + 41 = 5

Induction 5-25

CS1021 Divisibility problem (continued)'

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Step: Suppose n is such that f (n) is divisible by 5

We need to show that f (n + 1) is also divisible by 5

f (n + 1) = 32(n+1) + 4(n+1)+1

= 32n+2 + 4n+2

= 32n.32 + 4n+1.4

= 32n.9 + 4n+1.4

= 32n(5 + 4)+ 4n+1.4

= 5.32n + 4(32n + 4n+1)

Now 32n + 4n+1 is divisible by 5 (by inductive assumption), and so is 5.32n,

therefore so is f (n + 1), which completes the proof.

Induction 5-26

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Recursive function problem

f : N > Z

satisfies

f (0) = 0

f (1) = 1

f (r + 2) = 2 f (r + 1)− f (r)+ 2

∀r ∈ N.

We want to show that

∀r ∈ N, f (r) = r2

holds.

Induction 5-27

CS1021 Recursive function problem (continued)'

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We use induction on r.

Base (r = 0) Need to show that f (0) = 02

(This is trivial.)

Step (r 7→ r + 1)

Need to show that if f (r) = r2 then f (r + 1) = (r + 1)2.

This induction hypothesis doesn’t work

Induction 5-28


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We show the following.

∀r ∈ N, both

f (r) = r2 , f (r + 1) = (r + 1)2

hold.

Base (r = 0) Need to show that f (0) = 02 , f (1) = 12

(This is trivial.)

Step (r 7→ r + 1)

Need to show that if

f (r) = r2 , f (r + 1) = (r + 1)2

then

f (r + 1) = (r + 1)2 , f (r + 2)∗= (r + 2)2.

We derive the second conclusion ∗ from both hypotheses.

Induction 5-29


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f (r + 2)

= 2 f (r + 1)− f (r)+ 2 from the spec

= 2(r + 1)2− r2 + 2 by I. H.

= r2 + 4r + 4 arithmetic

= (r + 2)2 arithmetic

Induction 5-30

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A currency example

UK currency has coins of denominations 1p, 2p, 5p, 10p, 20p, 50p

Show that, if the 1p coin was abolished, the only quantities no longer

obtainable would be 1p and 3p.

Since 10p, 20p and 50p are all multiples of 5p we can clearly ignore them

and consider only 2p and 5p.

Induction 5-31

CS1021 A currency example (continued)'

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Without the 1p coin we can not obtain 1p and 3p (obvious).

We now need to show that for any amount n (≥ 4), we can make up n using

only 2p and 5p coins.

Base: (n=4)

4 = 2 + 2, so this is ok

Induction 5-32

CS1021 A currency example (continued)'

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Step: Suppose n is obtainable using 2p and 5p coins only. We will show that

(n + 1) is also obtainable.

2 possible cases. Since n≥ 4, we must use either

i) At least 2 2p coins, or

ii) at least 1 5 p coin

In case i), we can replace 2×2 by a single 5p coin to make n + 1.

In case ii), we can replace a single 5p coin by 3×2 to make n + 1.

Induction 5-33

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Structural Induction

The reason induction works on the natural numbers is because of their

recursive structure. Every natural number is either

• zero

• the successor of another natural number

Other structures are also built recursively and their structure can be used to

provide an induction mechanism

Induction 5-34

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Induction on Formal Languages

Recall that we define a formal language by giving a set of rules for producing

the words or expressions of the language.

These definitions are almost always recursive, and so induction techniques

similar to those above can be used to prove properties of the language

Induction 5-35

CS1021 Induction on Formal Languages (continued)'

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Suppose we define a formal language L of strings of 0’s and 1’s as follows

1. The empty string is in L

2. If a string s is in L , then so is the string 1s. The set of strings generated

by this rule is called M .

3. If a string s is in M , then 0s is in L

4. All strings in L are generated by the above rules.

The following strings are all members of L .

1

11

111

0111

10111

Induction 5-36


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100111 is not a member of L . Why?

Prove that, for all members s of L , s contains at least as many 1’s as 0’s, and

if s is a member of M it contains more 1’s than 0’s.

Call this property P(s)

Induction 5-37


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1. Base case: Empty string, ok

2. If s satisfies P(s), then clearly, so does 1s. By adding an extra 1 we

ensure that the string 1s contains more 1’s than 0’s.

3. Suppose s is in M , then the string 0s contains at least as many 1’s as

0’s (s had more 1’s than 0’s)

By induction, P(s) is true ∀s ∈ L

Induction 5-38

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5: Induction - School of Computer Science, University of ...graham/cs1021/slides/induction.pdf · The Principle of Induction Induction 5-1. CS1021 The natural numbers (continued)