1 Inverses and GCDs Supplementary Notes Prepared by Raymond Wong Presented by Raymond Wong

1

Inverses and GCDs

Supplementary Notes

Prepared by Raymond WongPresented by Raymond Wong

2

e.g.1 (Page 4)

E.g., 30 can be expressed as 1 x 2 x 3 x 5

2 divides 30 2 | 30

3 divides 30 3 | 30

5 divides 30 5 | 30

6 divides 30 6 | 30

10 divides 30 10 | 30

15 divides 30 15 | 30

30 divides 30 30 | 30

7 does not divide 301 divides 30 1 | 30

7 | 30

composite

3

e.g.2 (Page 4)

E.g., 24 can be expressed as 1 x 2 x 2 x 2 x 3

2 divides 24 2 | 24

3 divides 24 3 | 24

4 divides 24 4 | 24

6 divides 24 6 | 24

8 divides 24 8 | 24

12 divides 24 12 | 24


7 | 24

24 divides 24 24 | 24

composite

4

e.g.3 (Page 4)

E.g., 11 can be expressed as 1 x 11

11 divides 11 11 | 11


7 | 11

prime

5

e.g.4 (Page 4)

E.g., Is the following correct?7 | 0

0 can be expressed as

0 x 7

6

e.g.5 (Page 5)

E.g., What is gcd(7, 0)?

7 | 7and 7 | 0

E.g., Let n be a non-negative integer. What is gcd(n, 0)?

n | nand n | 0

7

e.g.6 (Page 7) Illustration of Theorem 2.15 E.g., j = 27

k = 5827 and 58 are relatively prime (i.e., gcd(27, 58) = 1)

if

then there exists two integers x and y such that 27x + 58y = 1

x = -15y = 7

27 and 58 are relatively prime (i.e., gcd(27, 58) = 1)

if

then

there exists two integers x and y such that 27x + 58y = 1

8

e.g.6 (Page 7) Illustration of Corollary 2.16 E.g. a = 27

n = 5827 has a multiplicative inverse (with respect to 58)

if

then gcd(27, 58) = 1

27 has a multiplicative inverse (with respect to 58)

if

then

gcd(27, 58) = 1

9

e.g.7 (Page 10)

E.g., m = 21 n = 9 21 can be expressed as 9 x 2 + 3 (i.e., nq + r)

q = 2r = 3

0 r < n

r is defined to be 21 mod 9

21 mod 9 is equal to 3

10

e.g.8 (Page 11)

Illustration of “Proof by Contradiction”We are going to prove that a claim C is correct

Proof by Contradiction:

Suppose “NOT C”

….

Derive some results, which may contradict to 1. “NOT C”, OR

2. some facts

e.g., we derived that C is true finally

e.g., we derived that “1 = 4”

11

e.g.9 (Page 11)

Illustration of “Proof by smallest counter example”

We are going to prove the following claim C: statement P(m) is true for each non-negative integer m, namely 0, 1, 2, …

P(0) true

P(1) true

P(2) true

P(3) true

P(4) true

If we can prove that statement P(m) is true for each non-negative integer separately, then we can prove the above claim C is correct.

Suppose that I want to prove that the above claimis correct by “Proof by Contradiction”.

… true

12

e.g.9



P(0) true

P(1) true

P(2) true

P(3) true

P(4) true

We can assume that there exists a non-negative integer k’ such that P(k’) is false


… true

false

Suppose “NOT C”.

There may exist another non-negative integer k such that P(k) is false

false

13

e.g.9



P(0) true

P(1) true

P(2) true

P(3) true

P(4) true


… true

false

Suppose “NOT C”.

false

We can assume that there exists a smallest non-negative integer k such that P(k) is false Why?

This is called by “Proof by smallest counter example”.

14

e.g.10 (Page 11)

We want to prove the following theorem.Theorem 2.12 (Euclid’s Division Theorem): Let n be a positive integer.

For every nonnegative integer m, there exist unique integers q, r such that m = nq + r and 0 r < n

15

e.g.10



Claim 1: For every nonnegative integer m, there exist integers q, r such that m = nq + r and 0 r < n

Claim 2: This pair q, r is unique.

16

e.g.10


17

e.g.10


P(m)

Claim CSuppose that there exists an integer m that P(m) is false

Proof by contradiction.

Proof by smallest counter example.

Suppose that there exists a “smallest” integer m that P(m) is false

There do not exist integers q, r such that m = nq + r and 0 r < n

Consider two cases.

Case 1: m < n Case 2: m n

We can write m = 0 + m

= n.0 + m

= nq + r where q = 0 and r = m

We conclude that there exist integers q, r such that m = nq + r and 0 r < n Contradiction

18

e.g.10


P(m)






Consider two cases.

Case 2: m n

19

e.g.10


P(m)






Consider two cases.

Case 2: m nWe know that m-n 0

Thus, m-n is a non-negative integer.

Since m-n is smaller than m,

there exist integers q’, r’ such that m-n = nq’ + r’ and 0 r’ < n

Consider m-n = nq’ + r’

m = nq’ + n + r’= n(q’ + 1) + r’

= nq + r

where q = q’+1 and r = r’

We conclude that there exist integers q, r such that m = nq + r and 0 r < n Contradiction

20

e.g.10


P(m)






Consider two cases.

In both cases, there are contradictions.

This implies that Claim 1 is correct.

21

e.g.10





22

e.g.10



23

e.g.10




Suppose that this pair q, r is not unique.

There exists a pair (q, r) and another pair (q’, r’) (where (q, r) (q’, r’)) such that m = nq + r …(*) and 0 r < n and m = nq’ + r’ …(**) and 0 r’ < n

Consider (*) – (**)

m - m = (nq+r) – (nq’ + r’)

0 = nq+r – nq’ - r’

0 = n(q-q’)+(r - r’)

n(q-q’)= r’ - r

r’ - r = n(q-q’)

Consider r’ - r

What is the greatest possible value?

< n - r n - 0

= n

r’ – r < n

Consider r’ - r

What is the smallest possible value?

> r’ - n

0 - n

= -n

r’ – r > -n

We conclude that |r’ – r| < n

-(r’ – r) < n

24

e.g.10




Suppose that this pair q, r is not unique.

There exists a pair (q, r) and another pair (q’, r’) (where (q, r) (q’, r’)) such that m = nq + r …(*) and 0 r < n and m = nq’ + r’ …(**) and 0 r’ < n

Consider (*) – (**)

m - m = (nq+r) – (nq’ + r’)

0 = nq+r – nq’ - r’

0 = n(q-q’)+(r - r’)

n(q-q’)= r’ - r

r’ - r = n(q-q’)

We conclude that |r’ – r| < n

We conclude that |r’ – r| < n |n(q-q’)| < n

We conclude that q – q’ = 0 q = q’Note that n(q-q’)= r’ - r 0 = r’ –

rr = r’

integer

We conclude that q = q’ and r = r’ (i.e., (q, r) = (q’, r’))

Contradiction

25

e.g.11 (Page 17)

Illustration of Lemma 2.13

Consider two integers 102 and 70.

Suppose that we can write 102 as 102 = 70.1 + 32

k = 102j = 70

q = 1r = 32

According to the lemma, we have gcd(102, 70) = gcd(70, 32)

26

e.g.12 (Page 17)

Prove the following lemma is correct.If j, k, q and r are non-negative integers such that

k = jq + rthen gcd(j, k) = gcd(r, j)

27

e.g.12

If j, k, q and r are non-negative integers such that k = jq + rthen gcd(j, k) = gcd(r, j)

Consider two cases.

Case 1: r = 0 Case 2: r > 0

Since k = jq + r,

we have k = jq

Consider gcd(j, k) = j

e.g., if 10 = 2qthen gcd(2, 10) = 2

Consider gcd(r, j)= gcd(0, j)

= j

Thus, gcd(j, k) = gcd(r, j)

e.g., gcd(0, 7) = 7

28

e.g.12


Consider two cases.

Case 2: r > 0

29

e.g.12


Consider two cases.Case 2: r > 0

We want to prove the following.

Claim 1:If d is a common divisor of j and k,then d is a common divisor of r and j.

Claim 2:If d is a common divisor of r and j,then d is a common divisor of j and k.

30

e.g.12



Let d be a common divisor of j and k

j can be written as j = i1d where i1 is a non-negative integer

k can be written as k = i2d where i2 is a non-negative integer

Consider k = jq + r

r = k – jq

=i2d – i1d.q

=(i2 – i1q)d

We conclude that d is a divisor of r

d is a common divisor of r and j

d is a divisor of j

d is a divisor of k

Since d is a divisor of j


31

e.g.12



Let d be a common divisor of r and j

r can be written as r = i3d where i3 is a non-negative integer

j can be written as j = i1d where i1 is a non-negative integer

Consider k = jq + r

= i1d.q + i3d

= (i1q + i3)d

We conclude that d is a divisor of k

d is a common divisor of j and k

d is a divisor of r

d is a divisor of j

Since d is a divisor of j



32

e.g.12





From Claim 1 and Claim 2, we conclude that

d is a common divisor of j and k if and only ifd is a common divisor of r and j.

We conclude that gcd(j, k) = gcd(r, j)

d is not a common divisor of j and k if and only ifd is not a common divisor of r and j.

A set of common divisors of j and k

A set of common divisors of r and j

5

7

11 5

7

11

A set of non-common divisors of j and k

2

…

3 2

…

3 A set of non-common divisors of r and j

33

e.g.13 (Page 17) How to use Lemma 2.13 for Euclid’s GCD

algorithm

Consider two integers 102 and 70.

Suppose that we can write 102 as 102 = 70.1 + 32

k = 102J = 70

q = 1r = 32

According to the lemma, we have gcd(102, 70) = gcd(70, 32)

Note that 70 = 32.2 + 6

gcd(70, 32) = gcd(32, 6)

Suppose that we want to find gcd(102, 70)

We can use Lemma 2.13 to compute gcd(102, 70)

Note that 32 = 6.5 + 2

gcd(32, 6) = gcd(6, 2)

Note that 6 = 2.3 + 0 gcd(6, 2) = gcd(2, 0)

Thus, gcd(102, 70) = gcd(2, 0) = 2

This corresponds to r.r decreases and finally its value becomes 0.

34

e.g.13

70 = 32.2 + 6


32 = 6.5 + 2

6 = 2.3 + 0

102 = 70.1 + 32

k = j.q + r k j q r

102 70

1 32

70 32

2 6

32 6 5 2

6 2 3 0

gcd(102, 70) = gcd(2, 0) = 2

35

e.g.14 (Page 24)

Definition of Multiplicative Inverse Given a positive integer n,

we define Zn = {0, 1, 2, …, n-1}

Given a value a Zn, a is said to have a multiplicative inverse a’

in Zn if a’ .n a = 1

36

e.g.14

E.g., n = 9 Z9 = {0, 1, 2, …, 8} Does 2 have a multiplicative inverse in Z9?

We may try all possible values in Z9

0 .9 2 = 0 0 is not a multiplicative inverse of 2 in Z9





5 .9 2 = 1 5 is a multiplicative inverse of 2 in Z9




2 has a multiplicative inverse 5 in Z9.Yes

37

e.g.14

E.g., n = 9 Z9 = {0, 1, 2, …, 8} Does 3 have a multiplicative inverse in Z9?

We may try all possible values in Z9










3 does not have a multiplicative inverse in Z9.No

38

e.g.15 (Page 25) Illustration of Lemma 2.5

Suppose that we want to find a value x in Z9 such that 2 .

9 x = 3 ……………(*) If 2 has a multiplicative inverse 5 in Z9

then x = 5 .9 3

and this solution is unique.

Why is it correct?

2 .9 x =

35 .9 (2 .

9 x) = 5 .9 3

(5 .9 2) .

9 x = 5 .9 3

1 .9 x = 5 .

9 3 x = 5 .

9 3

The computation/derivation in the right-hand-side box is valid for any x that satisfies equation (*).Thus, we conclude that only x that satisfies the equation (*) is 5 .

9 3

Why is this solution unique?

39

e.g.16 (Page 26)

Illustration of Theorem 2.7If 2 has a multiplicative inverse 5 in Z9

then the inverse 5 is unique.

According to Lemma 2.5

If 2 has a multiplicative inverse 5 in Z9

then x = 5 .9 b

and this solution is unique.

Why is it correct?

Consider 2 .9 x = b ……(*)

If we set b = 1, the equation (*) becomes 2 .

9 x = 1

According to Lemma 2.5, we have x = 5 .

9 1 and this solution is unique.

According to the inverse definition, x is an inverse of 2

40

e.g.17 (Page 27)

Please find each non-zero value a Z5 such that a has a multiplicative inverse a’ in Z5. (i.e., a .

5 a’ = 1) For each non-zero a Z5 and each non-zero b Z5, we compute a .

5 b if the above answer = 1 then we know that a has a multiplicative inverse b in Z5

or b has a multiplicative inverse a in Z5

41

e.g.17

For each non-zero a Z5 and each non-zero b Z5, we compute a .



Z5 = {0, 1, 2, 3, 4}

a = 1 and b = 1

1 .5 1 = 1

a = 1 and b = 2

1 .5 2 = 2

a = 1 and b = 3

1 .5 3 = 3

a = 1 and b = 4

1 .5 4 = 4

a = 2 and b = 1

2 .5 1 = 2

a = 2 and b = 2

2 .5 2 = 4

a = 2 and b = 3

2 .5 3 = 1

a = 2 and b = 4

2 .5 4 = 3

a = 3 and b = 1

3 .5 1 = 3

a = 3 and b = 2

3 .5 2 = 1

a = 3 and b = 3

3 .5 3 = 4

a = 3 and b = 4

3 .5 4 = 2

a = 4 and b = 1

4 .5 1 = 4

a = 4 and b = 2

4 .5 2 = 3

a = 4 and b = 3

4 .5 3 = 2

a = 4 and b = 4

4 .5 4 = 1

42

e.g.17




Z5 = {0, 1, 2, 3, 4}

a = 1 and b = 1

1 .5 1 = 1

a = 1 and b = 2

1 .5 2 = 2

a = 1 and b = 3

1 .5 3 = 3

a = 1 and b = 4

1 .5 4 = 4

a = 2 and b = 1

2 .5 1 = 2

a = 2 and b = 2

2 .5 2 = 4

a = 2 and b = 3

2 .5 3 = 1

a = 2 and b = 4

2 .5 4 = 3

a = 3 and b = 1

3 .5 1 = 3

a = 3 and b = 2

3 .5 2 = 1

a = 3 and b = 3

3 .5 3 = 4

a = 3 and b = 4

3 .5 4 = 2

a = 4 and b = 1

4 .5 1 = 4

a = 4 and b = 2

4 .5 2 = 3

a = 4 and b = 3

4 .5 3 = 2

a = 4 and b = 4

4 .5 4 = 1

1 has a multiplicative inverse 1 in Z52 has a multiplicative inverse 3

in Z5

3 has a multiplicative inverse 2 in Z5


2 has a multiplicative inverse 3 in Z54 has a multiplicative inverse 4

in Z5

a 1 2 3 4Inverse 1 3 2 4

43

e.g.18 (Page 27)

Please find each non-zero value a Z6 such that a has a multiplicative inverse a’ in Z6. (i.e., a .

6 a’ = 1) For each non-zero a Z6 and each non-zero b Z6, we compute a .



44

e.g.18




Z6 = {0, 1, 2, 3, 4, 5}

a = 1 and b = 1

1 .6 1 = 1

a = 1 and b = 2

1 .6 2 = 2

a = 1 and b = 3

1 .6 3 = 3

a = 1 and b = 4

1 .6 4 = 4

a = 1 and b = 5

1 .6 5 = 5

a = 2 and b = 1

2 .6 1 = 2

a = 2 and b = 2

2 .6 2 = 4

a = 2 and b = 3

2 .6 3 = 0

a = 2 and b = 4

2 .6 4 = 2

a = 2 and b = 5

2 .6 5 = 4

a = 3 and b = 1

3 .6 1 = 3

a = 3 and b = 2

3 .6 2 = 0

a = 3 and b = 3

3 .6 3 = 3

a = 3 and b = 4

3 .6 4 = 0

a = 3 and b = 5

3 .6 5 = 3

a = 4 and b = 1

4 .6 1 = 4

a = 4 and b = 2

4 .6 2 = 2

a = 4 and b = 3

4 .6 3 = 0

a = 4 and b = 4

4 .6 4 = 4

a = 4 and b = 5

4 .6 5 = 2

a = 5 and b = 1

5 .6 1 = 5

a = 5 and b = 2

5 .6 2 = 4

a = 5 and b = 3

5 .6 3 = 3

a = 5 and b = 4

5 .6 4 = 2

a = 5 and b = 5

5 .6 5 = 1

45

e.g.18




Z6 = {0, 1, 2, 3, 4, 5}

a = 1 and b = 1

1 .6 1 = 1

a = 1 and b = 2

1 .6 2 = 2

a = 1 and b = 3

1 .6 3 = 3

a = 1 and b = 4

1 .6 4 = 4

a = 1 and b = 5

1 .6 5 = 5

a = 2 and b = 1

2 .6 1 = 2

a = 2 and b = 2

2 .6 2 = 4

a = 2 and b = 3

2 .6 3 = 0

a = 2 and b = 4

2 .6 4 = 2

a = 2 and b = 5

2 .6 5 = 4

a = 3 and b = 1

3 .6 1 = 3

a = 3 and b = 2

3 .6 2 = 0

a = 3 and b = 3

3 .6 3 = 3

a = 3 and b = 4

3 .6 4 = 0

a = 3 and b = 5

3 .6 5 = 3

a = 4 and b = 1

4 .6 1 = 4

a = 4 and b = 2

4 .6 2 = 2

a = 4 and b = 3

4 .6 3 = 0

a = 4 and b = 4

4 .6 4 = 4

a = 4 and b = 5

4 .6 5 = 2

a = 5 and b = 1

5 .6 1 = 5

a = 5 and b = 2

5 .6 2 = 4

a = 5 and b = 3

5 .6 3 = 3

a = 5 and b = 4

5 .6 4 = 2

a = 5 and b = 5

5 .6 5 = 1



a 1 2 3 4 5Inverse 1 5X X X

46

e.g.18

a 1 2 3 4

Multiplicative inverse

1 3 2 4Z5:

a 1 2 3 4 5


1 X X X 5Z6:

a 1 2 3 4 5 6


1 4 5 2 3 6Z7:

a 1 2 3 4 5 6 7


1 X 3 X 5 X 7Z8:

a 1 2 3 4 5 6 7 8


1 5 X 7 2 X 4 8Z9:

47

e.g.19 (Page 30)

Illustration of Corollary 2.6 If there is a b Z6 (e.g., 3) such that

2 .6 x = b ………… (*)

does not have a solution, then 2 does not have a multiplicative inverse in Z6

Why is it correct?Proof by contradiction

Suppose that 2 has a multiplicative inverse x’ in Z6

By Lemma 2.5, we know that equation “2 .6 x = b” has a solution x = x’ .

6 b

Lemma 2.5 Consider equation 2 .

6 x = bIf 2 has a multiplicative inverse x’ in Z6

equation “2 .6 x = b” has a solution x = x’ .

6 b

This leads to a contradiction that equation “2 .

6 x = b” does not have a solution.

The equation “2x mod 6 = 3” does not have a solution2x is equal to an even number.2x mod 6 is also equal to an even number.

48

e.g.19 Illustration of Corollary 2.6

If there is a b Z6 (e.g., 3) such that 2 .

6 x = b ………… (*)does not have a solution, then 2 does not have a multiplicative inverse in Z6

The equation “2x mod 6 = 3” does not have a solution2x is equal to an even number.2x mod 6 is also equal to an even number.

Consider that the exam question asks you whether 2 has a multiplicativeinverse in Z6.

How will we use this corollary?

Suppose that we find that the equation “2x mod 6 = 3” does not have a solution (i.e., 2 .

6 x = 3 does not have a solution)

According to this corollary, we conclude that 2 does not have a multiplicativeinverse in Z6.

a 1 2 3 4 5Inverse 1 5X X X

In some of our previous slides, wederive that 2 does not have a multiplicativeinverse in Z6 by checking the table.

Z6

49

e.g.20 (Page 36)

Illustration of Lemma 2.8The modular equation 2 .

7 x = 1 has a solution in Z7

if and only if there exist integers x, y such that 2x + 7y = 1

Suppose that we know the modular equation 2 .7 x = 1 has a solution x = 4

Only if

We know that there exist integers x, y such that 2x + 7y = 1(In this case, x = -3 and y = 1)

We know the modular equation 2 .7 x = 1 has a solution x = 4

IfSuppose that we know that there exist integers x, y such that 2x + 7y = 1(In this case, x = -3 and y = 1)

50

e.g.20




Why is it correct?

Only ifThe modular equation 2 .

7 x = 1 has a solution x in Z7

We can write as 2x mod 7 = 1

We can re-write as 2x = 7q + 1 where q is an integer2x – 7q = 1

2x + 7(-q) = 1

Thus, there exist integers x, y such that 2x + 7y = 1 where y = -q

51

e.g.20




Why is it correct?

ifThere exist integers x, y such that 2x + 7y = 1

2x = -7y + 1

2x = (-y)7 + 1

We can re-write 2x mod 7 = 1

We can re-write 2 .7 x = 1

Thus, the modular equation 2 .7 x = 1 has a solution in Z7

52

e.g.21 (Page 37)

Illustration of Lemma 2.8/Theorem 2.9Lemma 2.8

The modular equation 2 .7 x = 1 has a solution in Z7


Theorem 2.92 has a multiplicative inverse in Z7


The above lemma can be restated as follows.

53

e.g.21



54

e.g.21



This theorem can help us find the inverse.

Corollary 2.10If there exist integers x, y such that 2x + 7y = 1,then the multiplicative inverse of 2 in Z7 is x mod 7

We want to show that 2 .7 x = 1

If this is true, then the multiplicative inverse of 2 in Z7 is x mod 7.

Consider 2 .7 x = 2 . x mod 7

Why is it correct?

= (2 . x + 7y) mod 7

= (2x + 7y) mod 7= 1 mod 7

= 1

55

e.g.22 (Page 40) Illustration of Lemma 2.11

Lemma 2.11If there exist integers x, y such that 2x + 7y = 1,then gcd(2, 7) = 1 (i.e., 2 and 7 are relatively prime.)

Why is it correct?Let k be a common divisor of 2 and 7

2 can be written as 2 = sk where s is an integer

7 can be written as 7 = qk where q is an integerConsider 2x + 7y = 1

sk.x + qk.y = 1

k(sx + qy) = 1

k is an integer and the RHS is equal to 1k must be equal to 1 or -1

The only common divisors of 2 and 7 are 1 and -1Thus, gcd(2, 7) = 1

56

e.g.23 (Page 44)

70 = 32.2 + 6


32 = 6.5 + 2

6 = 2.3 + 0

102 = 70.1 + 32

k = j.q + r k j q r

102 70

1 32

70 32

2 6

32 6 5 2

6 2 3 0

gcd(102, 70) = gcd(2, 0) = 2

57

e.g.23

70 = 32.2 + 6


32 = 6.5 + 2

6 = 2.3 + 0

102 = 70.1 + 32

k = j.q + r k j q r

102 70

1 32

70 32

2 6

32 6 5 2

6 2 3 0

Suppose that we want to find two integers x, y such that 70x + 102y = gcd(102, 70)

gcd(102, 70) = gcd(2, 0) = 2

i

0

1

2

3

k[i] = j[i].q[i] + r[i] k[i] j[i] q[i] r[i] y[i] x[i]

0 1

58

e.g.23

70 = 32.2 + 6


32 = 6.5 + 2

6 = 2.3 + 0

102 = 70.1 + 32

k = j.q + r k j q r

102 70

1 32

70 32

2 6

32 6 5 2

6 2 3 0


gcd(102, 70) = gcd(2, 0) = 2

i

0

1

2

3


0 1

1 -5 0 – 5.1

59

e.g.23

70 = 32.2 + 6


32 = 6.5 + 2

6 = 2.3 + 0

102 = 70.1 + 32

k = j.q + r k j q r

102 70

1 32

70 32

2 6

32 6 5 2

6 2 3 0


gcd(102, 70) = gcd(2, 0) = 2

i

0

1

2

3


0 1

1 -5

1 – 2.(-5)

-5 11

60

e.g.23

70 = 32.2 + 6


32 = 6.5 + 2

6 = 2.3 + 0

102 = 70.1 + 32

k = j.q + r k j q r

102 70

1 32

70 32

2 6

32 6 5 2

6 2 3 0


gcd(102, 70) = gcd(2, 0) = 2

i

0

1

2

3


0 1

1 -5

-5 – 1.(11)

-5 11

11 -16

61

e.g.23

70 = 32.2 + 6


32 = 6.5 + 2

6 = 2.3 + 0

102 = 70.1 + 32

k = j.q + r k j q r

102 70

1 32

70 32

2 6

32 6 5 2

6 2 3 0


gcd(102, 70) = gcd(2, 0) = 2

i

0

1

2

3


0 1

1 -5

-5 11

11 -16

x = -16y = 11

70 (-16) + 102 (11) = 2

= gcd(102, 70)

Let us verify it!

This algorithm is called Euclid’s extended GCD algorithm.

Note that 70 (a smaller value)is multiplied by x (not y).

62

e.g.24 (Page 48)

Illustration of Theorem 2.14Theorem 2.14Given two integers 102, 70,Euclid’s extended GCD algorithm computes (1) gcd (102, 70), and(2) two integers x, y such that 70x + 102y = gcd(102, 70)

Why is it correct?

We have already proved it.

How about this?

63

70 = 32.2 + 6

32 = 6.5 + 2

6 = 2.3 + 0

102 = 70.1 + 32

k = j.q + r k j q r

102 70

1 32

70 32

2 6

32 6 5 2

6 2 3 0

gcd(102, 70) = gcd(2, 0) = 2

i

0

1

2

3


0 1

1 -5

-5 11

11 -16

e.g.24

gcd(70, 102) = 70x + 102y

gcd(32, 70) = 32x + 70y

gcd(6, 32) = 6x + 32y

gcd(2, 6) = 2x + 6y

Why is it correct?

We want to show that there exist two integersx’ and y’ such that gcd(2, 6) = 2x’ + 6y’

64

e.g.24

We want to show that there exist two integersx’ and y’ such that gcd(2, 6) = 2x’ + 6y’

Note that, by Euclid’s Division Theorem, we can write 6 = 2.3 + rwhere r is equal to 0

gcd(2, 6) = 2

We can re-write the above expression as follows.

gcd(2, 6) = 2.1 + 6.0= 2x’ + 6y’

where x’ = 1 and y’ = 0

This is reason why we need to set x’ = 1 and y’ = 0 in the Extended GCD Algorithm

65

70 = 32.2 + 6

32 = 6.5 + 2

6 = 2.3 + 0

102 = 70.1 + 32

k = j.q + r k j q r

102 70

1 32

70 32

2 6

32 6 5 2

6 2 3 0

gcd(102, 70) = gcd(2, 0) = 2

i

0

1

2

3


0 1

1 -5

-5 11

11 -16

e.g.24

gcd(70, 102) = 70x + 102y

gcd(32, 70) = 32x + 70y

gcd(6, 32) = 6x + 32y

gcd(2, 6) = 2x + 6y

Why is it correct?

We want to show that there exist two integersx and y such that gcd(6, 32) = 6x + 32y

gcd(2, 6) = 2x’ + 6y’

This is correct.

66

32 = 6.5 + 2

6 = 2.3 + 0

2

3

e.g.24

gcd(6, 32) = 6x + 32y

2x + 6y = gcd(2, 6)


gcd(2, 6) = 2x’ + 6y’

67

32 = 6.5 + 2

6 = 2.3 + 0

2

3

e.g.24

gcd(6, 32) = 6x + 32y


gcd(2, 6) = 2x’ + 6y’ We have already proved that this is correct.

Next, we want to prove this is also correct.

Consider gcd(6, 32)= gcd(2, 6)

= 2x’ + 6y’

Note that gcd(6, 32) = gcd(2, 6)

= (32 – 6.5) x’ + 6y’

= 32x’ – 6.5.x’ + 6y’

= 6y’ – 6.5.x’ + 32x’ = 6(y’ – 5.x’) + 32x’ = 6x + 32y

where x = y’ – 5x’ and y = x’

This is the step we used in the Extended GCD algorithm.

x = y’ – 5x’ and y = x’According to x’, y’ and 5, we canfind the exact values of x and y.

68

32 = 6.5 + 2

6 = 2.3 + 0

2

3

e.g.24

gcd(6, 32) = 6x + 32y


gcd(2, 6) = 2x’ + 6y’


69

32 = 6.5 + 26 = 2.3 + 0

2

3

gcd(6, 32) = 6x + 32y


gcd(2, 6) = 2x’ + 6y’


70

70 = 32.2 + 6

32 = 6.5 + 2

6 = 2.3 + 0

102 = 70.1 + 32

k = j.q + r k j q r

102 70

1 32

70 32

2 6

32 6 5 2

6 2 3 0

i

0

1

2

3


0 1

1 -5 0 – 5.1

32 = 6.5 + 26 = 2.3 + 0

2

3

gcd(6, 32) = 6x + 32y


gcd(2, 6) = 2x’ + 6y’


y’ x’

y xx = y’ – 5x’

y = x’

71

e.g.25(Page 48) Illustration of Theorem 2.15

Theorem 2.15Two positive integers 27, 58 have gcd(27, 58) = 1(and thus they are relatively prime) if and only if there are integers x, y such that 27x + 58y =1

Why is it correct?Only if

We know that two positive integers 27, 58 have gcd(27, 58) = 1(and thus they are relatively prime)

Theorem 2.14Given two integers 27, 58,Euclid’s extended GCD algorithm computes (1) gcd (27, 58), and(2) two integers x, y such that 27x + 58y = gcd(27, 58)

By Theorem 2.14, we know that there are integers x, y such that 27x + 58y = 1

72

e.g.25 Illustration of Theorem 2.15


Why is it correct?If

We know that there are integers x, y such that 27x + 58y = 1

Lemma 2.11If there exist integers x, y such that 27x + 58y = 1,then gcd(27, 58) = 1 (i.e., 27 and 58 are relatively prime.)

By Lemma 2.11, we know that gcd(27, 58) = 1

73

e.g.26 (Page 49)Corollary 2.16Consider a positive integer 7.2 has a multiplicative inverse in Z7 iff gcd(2, 7) = 1. Why is it

correct?Lemma 2.8 The modular equation 2 .




74

e.g.26Corollary 2.16Consider a positive integer 7.2 has a multiplicative inverse in Z7 iff gcd(2, 7) = 1. Why is it

correct?Lemma 2.8 The modular equation 2 .




2 has a multiplicative inverse in Z7

gcd(2, 7) = 1

75

e.g.26

a 1 2 3 4


1 3 2 4Z5:

Since gcd(3, 5) = 1, 3 has the multiplicative inverse in Z5

76

e.g.26

a 1 2 3 4


1 3 2 4Z5:

a 1 2 3 4 5


1 X X X 5Z6:

Since gcd(3, 6) = 2 1, 3 has no multiplicative inverse in Z6

77

e.g.26

a 1 2 3 4


1 3 2 4Z5:

a 1 2 3 4 5


1 X X X 5Z6:

a 1 2 3 4 5 6


1 4 5 2 3 6Z7:

a 1 2 3 4 5 6 7


1 X 3 X 5 X 7Z8:

a 1 2 3 4 5 6 7 8


1 5 X 7 2 X 4 8Z9:

78

e.g.27 (Page 49)Corollary 2.17Note that 7 is a prime number. Every nonzero a Z7 has a multiplicative inverse. Why is it

correct?Since 7 is a prime number, gcd(a, 7) = 1

Corollary 2.16Consider a positive integer 7.a has a multiplicative inverse in Z7 iff gcd(a, 7) = 1. By the above corollary, we conclude that

a has a multiplicative inverse.

We know the following corollary.

79

e.g.27

a 1 2 3 4


1 3 2 4Z5:

a 1 2 3 4 5


1 X X X 5Z6:

a 1 2 3 4 5 6


1 4 5 2 3 6Z7:

a 1 2 3 4 5 6 7


1 X 3 X 5 X 7Z8:

a 1 2 3 4 5 6 7 8


1 5 X 7 2 X 4 8Z9:

Since 5 is a prime number, every non-zero a Z5 has a multiplicative inverse.

Since 5 is a prime number, every non-zero a Z5 has a multiplicative inverse.

80

e.g.27 (Page 52)

Illustration of Corollary 2.18Corollary 2.18If 2 has a multiplicative inverse in Z7,we can compute it by running Euclid’s extended GCD algorithm to determine integers x, y so that 2x + 7y = 1The inverse of 2 in Z7 is equal to x mod 7

Corollary 2.10If there exist integers x, y such that 2x + 7y = 1,then the multiplicative inverse of 2 in Z7 is x mod 7

Why is it correct?

81

e.g.28 (Page 52)

2 = 1.2 + 0

We want to find the multiplicative inverse of 2 in Z7

7 = 2.3 + 1 7 2 3 1

2 1 2 0

gcd(2, 7) = gcd(1, 0) = 1

0

1

k = j.q + ri k[i] = j[i].q[i] + r[i] k j q rk[i] j[i] q[i] r[i] y[i] x[i]

0 1

1 -3

Consider two integers 2 and 7

This implies that there exists a multiplicative inverse of 2 in Z7

0-3.1

x = -3y = 1

The algorithm finds 2x +7y = 1 (i.e., 2(-3) + 7(1) = 1)

The multiplicative inverse of 2 in Z7 is -3 mod 7 = 4

Documents

1 Inverses and GCDs Supplementary Notes Prepared by Raymond Wong Presented by Raymond Wong