Discrete Mathematics with Applications MATH236...Euler ˚-function Mathematical Induction Proof. Let...

Discrete Mathematics with ApplicationsMATH236

Dr. Hung P. Tong-Viet

School of Mathematics, Statistics and Computer ScienceUniversity of KwaZulu-NatalPietermaritzburg Campus

Semester 1, 2013

Tong-Viet (UKZN) MATH236 Semester 1, 2013 1 / 38

Table of contents

1 Exponentiation in Zm

2 Prime numbers

3 Euler φ-function

4 Fermat and Euler Theorems


Exponentiation in Zm

Square and mutiply

Many cryptosystems requires you to be able to compute things like

21351563 mod 3172

Use square and multiply algorithm:

Write the exponent as sum of powers of 2 (binary)

1563 = 210 + 29 + 24 + 23 + 2 + 1

Hence

21351563 = 2135210+29+24+23+2+1

= 2135210 · 21352

9 · 213524 · 21352

3 · 21352 · 2135



Square and multiply

Now 2135210

= (21352)29

so

21351563 = (21352)29 · 21352

9 · 213524 · 21352

3 · 21352 · 2135

It is easy to compute 21352 mod 3172 which is 61

So

21351563 ≡ (61)29 · 21352

9 · 213524 · 21352

3 · 21352 · 2135

≡ (61 · 2135)29 · 21352

4 · 213523 · 21352 · 2135

We now calculate 61 · 2135 ≡ 183 (mod 3172)

Begin the process over again, firstly, rewrite the term

(61 · 2135)29 ≡ 1832

9(mod 3172) = (1832)2

8

Continue in this fashion until we’re found the answer.



Square and multiply

21351563 ≡ (1832)28 · 21352

4 · 213523 · 21352 · 2135 (mod 3172)

≡ (1769)28 · 21352

4 · 213523 · 21352 · 2135 (mod 3172)

≡ (17692)27 · 21352

4 · 213523 · 21352 · 2135 (mod 3172)

≡ (1769)27 · 21352

4 · 213523 · 21352 · 2135 (mod 3172)

≡ (17692)26 · 21352

4 · 213523 · 21352 · 2135 (mod 3172)

≡ (1769)26 · 21352

4 · 213523 · 21352 · 2135 (mod 3172)

≡ (17692)25 · 21352

4 · 213523 · 21352 · 2135 (mod 3172)

≡ (17692)24 · 21352

4 · 213523 · 21352 · 2135 (mod 3172)

≡ (1769 · 2135)24 · 21352

3 · 21352 · 2135 (mod 3172)

≡ (61)23 · 21352

3 · 21352 · 2135 (mod 3172)

≡ (61 · 2135)23 · 21352 · 2135 (mod 3172)



Square and multiply

21351563 ≡ (183)22 · 21352 · 2135 (mod 3172)

≡ (183 · 2135)22 · 2135 (mod 3172)

≡ 61 · 2135 (mod 3172)≡ 183 (mod 3172)

Hence21351563 ≡ 183 (mod 3172)


Prime numbers

Prime numbers

An integer p ≥ 2 is a prime if 1 and p are its only positive divisors

Every positive integer n can be written uniquely in the formn = pe1

1 pe22 · · · p

ekk ,

where pi are distinct prime numbers and ei are positive integers

Prime numbers play an important role throughout the rest of thiscourse, particularly when we consider public-key cryptography


Prime numbers

Prime numbers

Here are some prime numbers:

2 3 5 7 11 13 17 19 23 2931 37 41 43 47 53 59 61 67 7173 79 83 89 97 101 103 107 109 113

127 131 137 139 149 151 157 163 167 173179 181 191 193 197 199 211 223 227 229233 239 241 251 257 263 269 271 277 281283 293 307 311 313 317 331 337 347 349353 359 367 373 379 383 389 397 401 409419 421 431 433 439 443 449 457 461 463


Prime numbers

Prime numbers

Theorem

There are infinitely many prime numbers


Prime numbers

Prime numbers

Proof.

Suppose to the contrary that there are finitely many prime numbers

Suppose that there are k prime numbers: p1, p2, · · · , pk

Consider the number n = p1p2 · · · pk + 1

Clearly n > pi for all i = 1, 2, · · · , k so n 6= pi for all i

Since n is congruence to 1 modulo pi for all i = 1, 2, · · · , k , we seethat pi - n for all i

Thus n must be a prime number

However, this contradicts the assumption that there are only k suchprime numbers.


Prime numbers

Prime numbers

The prime counting function π(n) is the number of prime numbersless than or equal to n.

π(2) = 1, π(3) = π(4) = 2, π(5) = π(6) = 3,π(7) = π(8) = π(9) = π(10) = 4

Theorem (Prime Number Theorem)

For large n,

π(n) ∼ n

ln(n)


Euler φ-function

Euler φ-function

Definition

For a positive integer n, we denote by φ(n) the number of positive integersa for which 1 ≤ a ≤ n and gcd(a, n) = 1. The function φ is called theEuler φ-function.

Example

Since 1, 5, 7 and 11 are all the positive integers less than 12 and relativelyprime to 12, we have that φ(12) = 4.


Euler φ-function

Euler φ-function

Table: Euler φ-function

n relatively prime integers in [1, n] φ(n)

1 1 12 1 13 1, 2 24 1, 3 25 1, 2, 3, 4 46 1, 5 27 1, 2, 3, 4, 5, 6 68 1, 3, 5, 7 49 1, 2, 4, 5, 7, 8 6

10 1, 3, 7, 9 4


Euler φ-function

Euler φ-function

Theorem

If p is a prime, then φ(p) = p − 1


Euler φ-function

Euler φ-function

Proof.

Let p be a prime

Then p is relatively prime to each of the numbers 1, 2, · · · , p − 1

So φ(p) = p − 1


Euler φ-function

Euler φ-function

Theorem

If p is a prime and k is a positive integer, then

φ(pk) = pk − pk−1 = pk(1− 1

p)


Euler φ-function

Euler φ-function

Proof.

Let S = {1, 2, · · · , pk}Let A be a subset of S consisting of numbers which is not relativelyprime to pk

If a ∈ A, then a is a multiple of p. Thus A = {p, 2p, 3p, · · · , pk−1p}There are pk−1 such multiples, so |A| = pk−1

Clearly, S − A = A is the set of all positive integers in {1, 2, · · · , pk}which are relatively prime to pk

So

φ(pk) = |A| = |S | − |A| = pk − pk−1 = pk(1− 1

p)


Euler φ-function

Euler φ-function

Example

φ(243) = φ(35) = 35(1− 1

3) = 162

φ(1024) = φ(210) = 210(1− 1

2) = 29 = 512


Euler φ-function

Mathematical Induction

Theorem (Principle of mathematical induction)

Let P(n) be a statement about n ∈ N. If

1 P(1) is true, and

2 For every k ∈ N, if P(k) is true, then P(k + 1) is true.

Then P(n) is true for all n ∈ N.


Euler φ-function


Example

Show that 13n − 1 is always divisible by 12 for all n ∈ N


Euler φ-function


Proof.

Let P(n) be the statement ‘13n − 1 is divisible by 12’

For n = 1, we have 131 − 1 = 12 is divisible by 12, so P(1) is true

Suppose that P(k) is true for some integer k ≥ 1. Then

13k − 1 is divisible by 12

We will prove that P(k + 1) is true, i.e., 13k+1 − 1 is divisible by 12

We have 13k+1 − 1 = (13 · 13k − 13) + (13− 1) = 13(13k − 1) + 12

Since 13k − 1 is divisible by 12 by the induction hypothesis and 12 isdivisible by 12

We obtain that 13(13k − 1) + 12 is divisible by 12

Hence 13k+1 − 1 is divisible by 12 so P(k + 1) is true

Thus by mathematical induction, 13n − 1 is divisible by 12


Euler φ-function


Example

Prove that 2n ≥ n + 1 for all n ∈ N. (∗)

Proof.

We have 21 = 2 ≥ 1 + 1 so (∗) is true for n = 1

Suppose that (∗) is true for n = k ≥ 1. Then 2k ≥ k + 1

We will prove that 2k+1 ≥ (k + 1) + 1 = k + 2

We have 2k+1 = 2 · 2k ≥ 2(k + 1) by induction hypothesis

We also have that 2(k + 1) = 2k + 2 ≥ k + 2 since k ≥ 1

So 2k+1 ≥ k + 2. Hence (∗) is true for k + 1.

By mathematical induction, we deduce that (∗) is true for all n ∈ N.


Euler φ-function

Difference Rule

Lemma

Let A be a subset of a finite set S . Then |A| = |S − A| = |S | − |A|.

Proof.

Since S = A ∪ A and A ∩ A = ∅, we obtain that

|S | = |A ∪ A| = |A|+ |A|.

Hence |A| = |S | − |A|.


Euler φ-function

Euler φ-function

Lemma

For k ≥ 1, let p1, p2, · · · , pk be distinct primes and let n be a positiveinteger that is divisible by pi for all i = 1, 2, · · · , k. Then the number ofintegers between 1 and n that are multiples of at least one of the integersii , 1 ≤ i ≤ k , i

n − nk∏

i=1

(1− 1

pi).


Euler φ-function

Euler φ-function

Proof.

We prove the lemma by induction on k

When k = 1, the integers between 1 and n that are multiples of p1 are

p1, 2p1, 3p1, · · · , (n

p1)p1

Hence, the number of integers between 1 and n that are multiples ofp1 is

n

p1= n − n(1− 1

p1) = n − n

k∏i=1

(1− 1

pi).

This establishes the base case


Euler φ-function

Euler φ-function

Proof.

Assume that k ≥ 2 and the result is true for all positive integers thatare divisible by k − 1 distinct primes

Let n be a positive integer that is divisible by k distinct primesp1, p2, · · · , pk

We wish to count the set S of all integers between 1 and n that aremultiples of at least one of the integers pi , 1 ≤ i ≤ k

We will use the Addition Rule

Let X be the set of all integers between 1 and n that are multiples ofat least one of the primes pi , 1 ≤ i ≤ k − 1

Let Y be the set of all integers between 1 and n that are multiples ofpk but not of pi for any i where 1 ≤ i ≤ k − 1


Euler φ-function

Euler φ-function

Proof.

By the Addition Rule, we have |S | = |X |+ |Y |By the induction hypothesis, we have

|X | = n − nk−1∏i=1

(1− 1

pi)

We next determine |Y |The set of all integers between 1 and n that are multiples of pk are

pk , 2pk , 3pk , · · · , (n

pk)pk

Hence, the number of integers between 1 and n that are multiples ofpk is n

pk


Euler φ-function

Euler φ-function

Proof.

Since gcd(pi , pk) = 1 for all 1 ≤ i < k , the multiples of pk that arenot multiples of pi for any i where 1 ≤ i ≤ k − 1 are those whosecoefficients (between 1 and n

pk) are not multiples of pi for any i ,

where 1 ≤ i ≤ k − 1.

Since npk

is divisible by pi for 1 ≤ i ≤ k − 1,

by induction hypothesis, the number of integers between 1 and npk

that are multiples of at least one of the integers pi , 1 ≤ i ≤ k − 1, is

n

pk− n

pk

k−1∏i=1

(1− 1

pi)


Euler φ-function

Euler φ-function

Proof.

It follows that

|Y | =n

pk− (

n

pk− n

pk

k−1∏i=1

(1− 1

pi)) =

n

pk

k−1∏i=1

(1− 1

pi)

Hence

|S | = |X |+ |Y |= n − n

∏k−1i=1 (1− 1

pi) + n

pk

∏k−1i=1 (1− 1

pi)

= n − [n∏k−1

i=1 (1− 1pi

)](1− 1pk

)

= n − n∏k

i=1(1− 1pi

)


Euler φ-function

Euler φ-function

Theorem

If n =∏k

i=1 peii , where p1, p2, · · · , pk are distinct primes and e1, e2, · · · , ek

are positive integers, then

φ(n) = nk∏

i=1

(1− 1

pk)


Euler φ-function

Euler φ-function

Since n =∏k

i=1 peii , where p1, p2, · · · , pk are distinct primes, and

e1, e2, · · · , ek are positive integers

The only integers between 1 and n that are not relatively prime to nare those integers between 1 and n that are multiples of at least oneof the integers pi , 1 ≤ i ≤ k

By our previous lemma, the number of such integers is

n − nk∏

i=1

(1− 1

pi)

Hence, the result follows by applying the difference rule.


Euler φ-function

Euler φ-function

Example

φ(30) = φ(2 · 3 · 5) = 2 · 3 · 5(1− 12)(1− 1

3)(1− 15) = 8

φ(6) = 2 · 3(1− 12)(1− 1

3) = 2

φ(20) = 20(1− 12)(1− 1

5) and φ(9) = 9− 3 = 6

We have φ(180) = φ(22 · 32 · 5) = 22 · 32 · 5(1− 12)(1− 1

3)(1− 15) = 48


Euler φ-function

Euler φ-function

Theorem

If m and n are relatively prime integers and gcd(m, n) = 1, then

φ(mn) = φ(m)φ(n).


Euler φ-function

Euler φ-function

Proof.

Let m =∏k

i=1 peii and n =

∏k ′

i=1(p′i )e′i , where

p1, p2, · · · , pk , p′1, p′2, · · · , p′k are distinct primes since gcd(m, n) = 1

By the previous theorem, we have

φ(mn) = mn∏k

i=1(1− 1pi

)∏k ′

i=1(1− 1p′i

)

= m∏k

i=1(1− 1pi

)n∏k ′

i=1(1− 1p′i

)

= φ(m)φ(n)


Euler φ-function

Euler φ-function

Example

We have φ(248832) = φ(210 · 35) = φ(210)φ(35) = 162 · 512 = 82944


Fermat and Euler Theorems

Fermat Theorem

Theorem (Fermat)

If a is a positive integer and p is a prime number, then

ap ≡ a (mod p)



Fermat Theorem

Proof.

We will prove this theorem by induction on a

If a = 1, then ap = 1p ≡ 1 = a (mod p) as wanted

Assume that for some integer a ≥ 1, we have

ap ≡ a (mod p)

We will prove that (a + 1)p ≡ a + 1 (mod p)

Consider the binomial expansion of (a + 1)p



Fermat Theorem

Proof.

(a + 1)p =

p∑i=0

(p

i

)ai = 1 + pa + · · ·+ pap−1 + ap

Except the first and the last terms, every other term is divisible by p

So (a + 1)p ≡ ap + 1 (mod p)

By the induction hypothesis, we have ap ≡ a (mod p)

Thus(a + 1)p ≡ a + 1 (mod p)


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