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

2. e.g.1 (Page 4) E.g., 30 can be expressed as 1 x 2 x 3 x 5 composite
1 divides 30
1 | 30
6 divides 30
6 | 30
7 does not divide 30
7 | 30
2 divides 30
2 | 30
10 divides 30
10 | 30
3 divides 30
3 | 30
15 divides 30
15 | 30
5 divides 30
5 | 30
30 divides 30
30 | 30

3. e.g.2 (Page 4) E.g., 24 can be expressed as 1 x 2 x 2 x 2 x 3
composite
1 divides 24
1 | 24
4 divides 24
4 | 24
7 does not divide 24
7 | 24
2 divides 24
2 | 24
6 divides 24
6 | 24
3 divides 24
3 | 24
8 divides 24
8 | 24
12 divides 24
12 | 24
24 divides 24
24 | 24

4. e.g.3 (Page 4) E.g., 11 can be expressed as 1 x 11 prime 1 divides 11
1 | 11
7 does not divide 11
11 divides 11
11 | 11
7 | 11

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 | 7
and 7 | 0
E.g., Let n be a non-negative integer. What is gcd(n, 0)?
n | n
and n | 0

7. e.g.6 (Page 7) Illustration of Theorem 2.15 E.g., j = 27 k = 58 if
27 and 58 are relatively prime (i.e., gcd(27, 58) = 1)
then
there exists two integers x and y such that x + 58y = 1
x = -15
y = 7
there exists two integers x and y such that x + 58y = 1 if
then
27 and 58 are relatively prime (i.e., gcd(27, 58) = 1)

8. e.g.6 (Page 7) Illustration of Corollary 2.16 E.g. a = 27 n = 58 if
27 has a multiplicative inverse (with respect to 58)
then
gcd(27, 58) = 1 if
gcd(27, 58) = 1
then
27 has a multiplicative inverse (with respect to 58)

9. e.g.7 (Page 10)
E.g., m = n = can be expressed as x (i.e., nq + r)
r is defined to be 21 mod 9
q = 2
r = 3
21 mod 9 is equal to 3
0  r < n

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 “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”
Suppose that I want to prove that the above claim
is correct by “Proof by Contradiction”.
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
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.
P(1)
true
P(2)
true
P(3)
true
P(4)
true …
true

12. e.g.9 Illustration of “Proof by smallest counter example”
Suppose that I want to prove that the above claim
is correct by “Proof by Contradiction”.
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
Suppose “NOT C”.
We can assume that there exists a non-negative integer k’ such that
P(k’) is false
P(1)
true
P(2)
false
true
P(3)
true
There may exist another non-negative integer k such that
P(k) is false
P(4)
false
true …
true

13. e.g.9 Illustration of “Proof by smallest counter example”
Suppose that I want to prove that the above claim
is correct by “Proof by Contradiction”.
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
Suppose “NOT C”.
P(1)
true
We can assume that there exists a smallest non-negative integer k such that
P(k) is false
P(2)
true
false
P(3)
true
Why?
P(4)
false
true
This is called by “Proof by smallest counter example”. …
true

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

17. Claim 1: For every nonnegative integer m, there exist integers q, r such that m = nq + r
and 0  r < n
e.g.10
P(m)
Proof by contradiction.
Claim C
Suppose that there exists an integer m that P(m) is false
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. Claim 1: For every nonnegative integer m, there exist integers q, r such that m = nq + r
and 0  r < n
e.g.10
P(m)
Proof by contradiction.
Claim C
Suppose that there exists an integer m that P(m) is false
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 2: m  n

19. Claim 1: For every nonnegative integer m, there exist integers q, r such that m = nq + r
and 0  r < n
e.g.10
P(m)
Proof by contradiction.
Claim C
Suppose that there exists an integer m that P(m) is false
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.
Consider m-n = nq’ + r’
Case 2: m  n
m = nq’ + n + r’
We know that m-n  0
= n(q’ + 1) + r’
Thus, m-n is a non-negative integer.
= nq + r
Since m-n is smaller than m,
where q = q’+1 and r = r’
there exist integers q’, r’ such that m-n = nq’ + r’
and 0  r’ < n
We conclude that there exist integers q, r such that m = nq + r
and 0  r < n
Contradiction

20. Claim 1: For every nonnegative integer m, there exist integers q, r such that m = nq + r
and 0  r < n
e.g.10
P(m)
Proof by contradiction.
Claim C
Suppose that there exists an integer m that P(m) is false
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.
In both cases, there are contradictions.
This implies that Claim 1 is correct.

21. e.g.10 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
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.

22. 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.

23. Claim 1: For every nonnegative integer m, there exist integers q, r such that m = nq + r
and 0  r < n
e.g.10
Claim 2: This pair q, r is unique.
Proof by contradiction.
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
What is the greatest possible value?
Consider (*) – (**)
Consider r’ - r
< n - r
m - m = (nq+r) – (nq’ + r’)
 n - 0
r’ – r < n
0 = nq+r – nq’ - r’
= n
0 = n(q-q’)+(r - r’)
What is the smallest possible value?
r’ - r = n(q-q’)
Consider r’ - r
> r’ - n
r’ – r > -n
n(q-q’)= r’ - r
 0 - n
-(r’ – r) < n
We conclude that |r’ – r| < n
= -n

24. Claim 1: For every nonnegative integer m, there exist integers q, r such that m = nq + r
and 0  r < n
e.g.10
Claim 2: This pair q, r is unique.
Proof by contradiction.
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 (*) – (**)
We conclude that |r’ – r| < n
integer
|n(q-q’)| < n
m - m = (nq+r) – (nq’ + r’)
We conclude that q – q’ = 0
0 = nq+r – nq’ - r’
q = q’
0 = n(q-q’)+(r - r’)
Note that n(q-q’)= r’ - r
r’ - r = n(q-q’)
0 = r’ – r
n(q-q’)= r’ - r
r = r’
Contradiction
We conclude that q = q’ and r = r’ (i.e., (q, r) = (q’, r’))
We conclude that |r’ – r| < n

25. e.g.11 (Page 17) Illustration of Lemma 2.13 k = 102 j = 70
Consider two integers 102 and 70.
Suppose that we can write 102 as
102 =
q = 1
r = 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 + r
then
gcd(j, k) = gcd(r, j)

27. e.g.12 If j, k, q and r are non-negative integers such that k = jq + r
then
gcd(j, k) = gcd(r, j)
e.g.12
Consider two cases.
Case 1: r = 0
Case 2: r > 0
e.g., if 10 = 2q
then gcd(2, 10) = 2
Since k = jq + r,
we have k = jq
Consider gcd(j, k)
= j
e.g., gcd(0, 7) = 7
Consider gcd(r, j)
= gcd(0, j)
= j
Thus, gcd(j, k) = gcd(r, j)

28. e.g.12 If j, k, q and r are non-negative integers such that k = jq + r
then
gcd(j, k) = gcd(r, j)
e.g.12
Consider two cases.
Case 2: r > 0

29. e.g.12 If j, k, q and r are non-negative integers such that k = jq + r
then
gcd(j, k) = gcd(r, j)
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 If j, k, q and r are non-negative integers such that k = jq + r
then
gcd(j, k) = gcd(r, j)
e.g.12
Claim 1:
If d is a common divisor of j and k,
then d is a common divisor of r and j.
Consider two cases.
Case 2: r > 0
Let d be a common divisor of j and k
d is a divisor of j
j can be written as
j = i1d where i1 is a non-negative integer
d is a divisor of k
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
Since d is a divisor of j
d is a common divisor of r and j

31. e.g.12 If j, k, q and r are non-negative integers such that k = jq + r
then
gcd(j, k) = gcd(r, j)
e.g.12
Claim 1:
If d is a common divisor of j and k,
then d is a common divisor of r and j.
Consider two cases.
Claim 2:
If d is a common divisor of r and j,
then d is a common divisor of j and k.
Case 2: r > 0
Let d be a common divisor of r and j
d is a divisor of r
r can be written as
r = i3d where i3 is a non-negative integer
d is a divisor of j
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
Since d is a divisor of j
d is a common divisor of j and k

32. e.g.12 If j, k, q and r are non-negative integers such that k = jq + r
then
gcd(j, k) = gcd(r, j)
e.g.12
Claim 1:
If d is a common divisor of j and k,
then d is a common divisor of r and j.
Consider two cases.
Claim 2:
If d is a common divisor of r and j,
then d is a common divisor of j and k.
Case 2: r > 0
From Claim 1 and Claim 2, we conclude that
d is a common divisor of j and k if and only if
d is a common divisor of r and j.
d is not a common divisor of j and k if and only if
d is not a common divisor of r and j.
A set of common divisors of j and k 5 7 11 5 7 11
A set of common divisors of r and j 2 … 3 2 … 3
A set of non-common divisors of r and j
A set of non-common divisors of j and k
We conclude that gcd(j, k) = gcd(r, j)

33. e.g.13 (Page 17) How to use Lemma 2.13 for Euclid’s GCD algorithm
Suppose that we want to find gcd(102, 70)
k = 102
J = 70
We can use Lemma 2.13 to compute gcd(102, 70)
Consider two integers 102 and 70.
Suppose that we can write 102 as
102 =
q = 1
r = 32
This corresponds to r.
r decreases and finally its value becomes 0.
According to the lemma, we have gcd(102, 70) = gcd(70, 32)
Note that =
gcd(70, 32) = gcd(32, 6)
Note that =
gcd(32, 6) = gcd(6, 2)
Note that 6 =
gcd(6, 2) = gcd(2, 0)
Thus, gcd(102, 70) = gcd(2, 0) = 2

34. e.g.13 Suppose that we want to find gcd(102, 70) k = j.q + r k j q r
102 =
102 70 1 32
70 = 70 32 2 6
32 = 32 6 5 2 6 2 3
6 =
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 = Z9 = {0, 1, 2, …, 8} Does 2 have a multiplicative inverse in Z9?
We may try all possible values in Z9
= 0
0 is not a multiplicative inverse of 2 in Z9
= 2
1 is not a multiplicative inverse of 2 in Z9
= 4
2 is not a multiplicative inverse of 2 in Z9
= 6
3 is not a multiplicative inverse of 2 in Z9
= 8
4 is not a multiplicative inverse of 2 in Z9
= 1
5 is a multiplicative inverse of 2 in Z9
= 3
6 is not a multiplicative inverse of 2 in Z9
= 5
7 is not a multiplicative inverse of 2 in Z9
= 7
8 is not a multiplicative inverse of 2 in Z9
Yes
2 has a multiplicative inverse 5 in Z9.

37. e.g.14
E.g., n = Z9 = {0, 1, 2, …, 8} Does 3 have a multiplicative inverse in Z9?
We may try all possible values in Z9
= 0
0 is not a multiplicative inverse of 3 in Z9
= 3
1 is not a multiplicative inverse of 3 in Z9
= 6
2 is not a multiplicative inverse of 3 in Z9
= 0
3 is not a multiplicative inverse of 3 in Z9
= 3
4 is not a multiplicative inverse of 3 in Z9
= 6
5 is not a multiplicative inverse of 3 in Z9
= 0
6 is not a multiplicative inverse of 3 in Z9
= 3
7 is not a multiplicative inverse of 3 in Z9
= 6
8 is not a multiplicative inverse of 3 in Z9 No
3 does not have a multiplicative inverse in Z9.

38. e.g.15 (Page 25) Illustration of Lemma 2.5
Suppose that we want to find a value x in Z9 such that x = ……………(*)
If 2 has a multiplicative inverse 5 in Z9
Why is it correct?
then x = and this solution is unique.
2 .9 x = 3
Why is this solution unique?
5 .9 (2 .9 x) =
The computation/derivation in the right-hand-side box is valid for any x that satisfies equation (*).
(5 .9 2) .9 x =
1 .9 x =
Thus, we conclude that only x that satisfies the equation (*) is
x =

39. e.g.16 (Page 26) Illustration of Theorem 2.7
If 2 has a multiplicative inverse 5 in Z9
then the inverse 5 is unique.
Why is it correct?
According to Lemma 2.5
Consider 2 .9 x = b ……(*)
If 2 has a multiplicative inverse 5 in Z9
then x = 5 .9 b and this solution is unique.
If we set b = 1, the equation (*) becomes
2 .9 x = 1
According to the inverse definition, x is an inverse of 2
According to Lemma 2.5, we have x = and this solution is unique.

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 = then we know that a has a multiplicative inverse b in Z or b has a multiplicative inverse a in Z5

41. For each non-zero a  Z5 and each non-zero b  Z5, we compute a .5 b
e.g.17
if the above answer = then we know that a has a multiplicative inverse b in Z or b has a multiplicative inverse a in Z5
Z5 = {0, 1, 2, 3, 4}
a = 1 and b = 1
a = 1 and b = 2
a = 1 and b = 3
a = 1 and b = 4
= 1
= 2
= 3
= 4
a = 2 and b = 1
a = 2 and b = 2
a = 2 and b = 3
a = 2 and b = 4
= 2
= 4
= 1
= 3
a = 3 and b = 1
= 3
a = 3 and b = 2
= 1
a = 3 and b = 3
= 4
a = 3 and b = 4
= 2
a = 4 and b = 1
= 4
a = 4 and b = 2
= 3
a = 4 and b = 3
= 2
a = 4 and b = 4
= 1

42. For each non-zero a  Z5 and each non-zero b  Z5, we compute a .5 b
e.g.17 a 1 2 3 4
Inverse
if the above answer = then we know that a has a multiplicative inverse b in Z or b has a multiplicative inverse a in Z5 1 3 2 4
Z5 = {0, 1, 2, 3, 4}
1 has a multiplicative inverse 1 in Z5
a = 1 and b = 1
a = 1 and b = 2
a = 1 and b = 3
a = 1 and b = 4
2 has a multiplicative inverse 3 in Z5
3 has a multiplicative inverse 2 in Z5
= 1
= 2
= 3
= 4
a = 2 and b = 1
a = 2 and b = 2
a = 2 and b = 3
a = 2 and b = 4
= 2
= 4
= 1
= 3
3 has a multiplicative inverse 2 in Z5
2 has a multiplicative inverse 3 in Z5
a = 3 and b = 1
= 3
a = 3 and b = 2
= 1
a = 3 and b = 3
= 4
a = 3 and b = 4
= 2
4 has a multiplicative inverse 4 in Z5
a = 4 and b = 1
= 4
a = 4 and b = 2
= 3
a = 4 and b = 3
= 2
a = 4 and b = 4
= 1

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 .6 b
if the above answer = then we know that a has a multiplicative inverse b in Z or b has a multiplicative inverse a in Z6

44. For each non-zero a  Z6 and each non-zero b  Z6, we compute a .6 b
e.g.18
if the above answer = then we know that a has a multiplicative inverse b in Z or b has a multiplicative inverse a in Z6
Z6 = {0, 1, 2, 3, 4, 5}
a = 1 and b = 1
= 1
a = 1 and b = 2
= 2
a = 1 and b = 3
= 3
a = 1 and b = 4
= 4
a = 1 and b = 5
= 5
a = 2 and b = 1
= 2
a = 2 and b = 2
= 4
a = 2 and b = 3
= 0
a = 2 and b = 4
= 2
a = 2 and b = 5
= 4
a = 3 and b = 1
= 3
a = 3 and b = 2
= 0
a = 3 and b = 3
= 3
a = 3 and b = 4
= 0
a = 3 and b = 5
= 3
a = 4 and b = 1
= 4
a = 4 and b = 2
= 2
a = 4 and b = 3
= 0
a = 4 and b = 4
= 4
a = 4 and b = 5
= 2
a = 5 and b = 1
= 5
a = 5 and b = 2
= 4
a = 5 and b = 3
= 3
a = 5 and b = 4
= 2
a = 5 and b = 5
= 1

45. For each non-zero a  Z6 and each non-zero b  Z6, we compute a .6 b
e.g.18 a 1 2 3 4 5
Inverse
if the above answer = then we know that a has a multiplicative inverse b in Z or b has a multiplicative inverse a in Z6 1 X X X 5
Z6 = {0, 1, 2, 3, 4, 5}
a = 1 and b = 1
= 1
a = 1 and b = 2
= 2
a = 1 and b = 3
= 3
a = 1 and b = 4
= 4
a = 1 and b = 5
= 5
1 has a multiplicative inverse 1 in Z6
a = 2 and b = 1
= 2
a = 2 and b = 2
= 4
a = 2 and b = 3
= 0
a = 2 and b = 4
= 2
a = 2 and b = 5
= 4
a = 3 and b = 1
= 3
a = 3 and b = 2
= 0
a = 3 and b = 3
= 3
a = 3 and b = 4
= 0
a = 3 and b = 5
= 3
a = 4 and b = 1
= 4
a = 4 and b = 2
= 2
a = 4 and b = 3
= 0
a = 4 and b = 4
= 4
a = 4 and b = 5
= 2
5 has a multiplicative inverse 5 in Z6
a = 5 and b = 1
= 5
a = 5 and b = 2
= 4
a = 5 and b = 3
= 3
a = 5 and b = 4
= 2
a = 5 and b = 5
= 1

46. e.g.18 a 1 2 3 4 Multiplicative inverse Z5: a 1 2 3 4 5X
Z6: a 1 2 3 4 5 6
Multiplicative inverse
Z7: a 1 2 3 4 5 6 7
Multiplicative inverse X
Z8: a 1 2 3 4 5 6 7 8
Multiplicative inverse X
Z9:

47. e.g.19 (Page 30) Illustration of Corollary 2.6
Lemma 2.5 Consider equation
2 .6 x = b
If 2 has a multiplicative inverse x’ in Z6
equation “2 .6 x = b” has a solution x = x’ .6 b
e.g.19 (Page 30)
The equation “2x mod 6 = 3” does not have a solution
Illustration of Corollary 2.6
If there is a b  Z6 (e.g., 3) such that x = b ………… (*) does not have a solution, then 2 does not have a multiplicative inverse in Z6
2x is equal to an even number.
2x mod 6 is also equal to an even number.
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
This leads to a contradiction that equation “2 .6 x = b” does not have a solution.

48. e.g.19 Illustration of Corollary 2.6 a 1 2 3 4 5
In some of our previous slides, we
derive that 2 does not have a multiplicative
inverse in Z6 by checking the table. Z6
e.g.19 a 1 2 3 4 5
Inverse 1 5 X
The equation “2x mod 6 = 3” does not have a solution
Illustration of Corollary 2.6
If there is a b  Z6 (e.g., 3) such that x = b ………… (*) does not have a solution, then 2 does not have a multiplicative inverse in Z6
2x is equal to an even number.
2x mod 6 is also equal to an even number.
How will we use this corollary?
Consider that the exam question asks you whether 2 has a multiplicative
inverse in Z6.
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 multiplicative inverse in Z6.

49. e.g.20 (Page 36) Illustration of Lemma 2.8
The 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
Only if
Suppose that we know the modular equation 2 .7 x = 1 has a solution x = 4
We know that there exist integers x, y such that 2x + 7y = 1 (In this case, x = -3 and y = 1) If
Suppose that 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

50. e.g.20 Illustration of Lemma 2.8
The 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
Why is it correct?
Only if
The modular equation 2 .7 x = 1 has a solution x in Z7
We can write as x mod 7 = 1
We can re-write as 2x = 7q + 1 where q is an integer
2x – 7q = 1
2x + 7(-q) = 1
Thus, there exist integers x, y such that 2x + 7y = 1
where y = -q

51. e.g.20 Illustration of Lemma 2.8
The 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
Why is it correct? if
There exist integers x, y such that 2x + 7y = 1
2x = -7y + 1
2x = (-y)7 + 1
We can re-write x mod 7 = 1
We can re-write 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 2.9
Lemma 2.8 The 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
The above lemma can be restated as follows.
Theorem 2.9
2 has a multiplicative inverse in Z7
if and only if
there exist integers x, y such that 2x + 7y = 1

53. e.g.21 Theorem 2.9 2 has a multiplicative inverse in Z7 if and only if
there exist integers x, y such that 2x + 7y = 1

54. e.g.21 Theorem 2.9 2 has a multiplicative inverse in Z7 if and only if
there exist integers x, y such that 2x + 7y = 1
e.g.21
This theorem can help us find the inverse.
Corollary 2.10
If 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?
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
= (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 2.11
If 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 integer
Consider 2x + 7y = 1
The only common divisors of 2 and 7 are 1 and -1
sk.x + qk.y = 1
k(sx + qy) = 1
Thus, gcd(2, 7) = 1
k is an integer and the RHS is equal to 1
k must be equal to 1 or -1

56. e.g.23 (Page 44) Suppose that we want to find gcd(102, 70) k = j.q + r
102 =
102 70 1 32 2 6 5 3
70 =
32 =
6 =
gcd(102, 70) = gcd(2, 0) = 2

57. e.g.23 Suppose that we want to find gcd(102, 70)
Suppose that we want to find two integers x, y such that x + 102y = gcd(102, 70) i
k[i] = j[i].q[i] + r[i]
k = j.q + r
k[i] k j q r
j[i]
q[i]
r[i]
y[i]
x[i]
102 =
102 70 1 32 2 6 5 3 1
70 = 2
32 = 1 3
6 =
gcd(102, 70) = gcd(2, 0) = 2

58. e.g.23 Suppose that we want to find gcd(102, 70)
Suppose that we want to find two integers x, y such that x + 102y = gcd(102, 70) i
k[i] = j[i].q[i] + r[i]
k = j.q + r
k[i] k j q r
j[i]
q[i]
r[i]
y[i]
x[i]
102 =
102 70 1 32 2 6 5 3 1
70 = 2
32 = 1 -5
0 – 5.1 1 3
6 =
gcd(102, 70) = gcd(2, 0) = 2

59. e.g.23 Suppose that we want to find gcd(102, 70)
Suppose that we want to find two integers x, y such that x + 102y = gcd(102, 70) i
k[i] = j[i].q[i] + r[i]
k = j.q + r
k[i] k j q r
j[i]
q[i]
r[i]
y[i]
x[i]
102 =
102 70 1 32 2 6 5 3 1
70 = -5 11
1 – 2.(-5) 2
32 = 1 -5 1 3
6 =
gcd(102, 70) = gcd(2, 0) = 2

60. e.g.23 Suppose that we want to find gcd(102, 70)
Suppose that we want to find two integers x, y such that x + 102y = gcd(102, 70) i
k[i] = j[i].q[i] + r[i]
k = j.q + r
k[i] k j q r
j[i]
q[i]
r[i]
y[i]
x[i]
-5 – 1.(11)
102 =
102 70 1 32 2 6 5 3 11
-16 1
70 = -5 11 2
32 = 1 -5 1 3
6 =
gcd(102, 70) = gcd(2, 0) = 2

61. e.g.23 This algorithm is called Euclid’s extended GCD algorithm.
y = 11
Note that 70 (a smaller value) is multiplied by x (not y).
Let us verify it!
70 (-16) (11) = 2
Suppose that we want to find gcd(102, 70)
= gcd(102, 70)
Suppose that we want to find two integers x, y such that x + 102y = gcd(102, 70) i
k[i] = j[i].q[i] + r[i]
k = j.q + r
k[i] k j q r
j[i]
q[i]
r[i]
y[i]
x[i]
102 =
102 70 1 32 2 6 5 3 11
-16 1
70 = -5 11 2
32 = 1 -5 1 3
6 =
gcd(102, 70) = gcd(2, 0) = 2

62. e.g.24 (Page 48) Illustration of Theorem 2.14
Theorem 2.14 Given two integers 102, 70,
Euclid’s extended GCD algorithm computes (1) gcd (102, 70), and (2) two integers x, y such that x + 102y = gcd(102, 70)
We have already proved it.
How about this?
Why is it correct?

63. e.g.24 We want to show that there exist two integers
x’ and y’ such that gcd(2, 6) = 2x’ + 6y’
e.g.24 i
k[i] = j[i].q[i] + r[i]
k = j.q + r
k[i] k j q r
j[i]
q[i]
r[i]
y[i]
x[i]
gcd(70, 102) = 70x + 102y
102 =
102 70 1 32 2 6 5 3 11
-16
gcd(32, 70) = 32x + 70y 1
70 = -5 11
gcd(6, 32) = 6x + 32y 2
32 = 1 -5
gcd(2, 6) = 2x + 6y 1 3
6 =
Why is it correct?
gcd(102, 70) = gcd(2, 0) = 2

64. e.g.24 We want to show that there exist two integers
x’ and y’ such that gcd(2, 6) = 2x’ + 6y’
e.g.24
Note that, by Euclid’s Division Theorem, we can write = r
where r is equal to 0
gcd(2, 6) = 2
We can re-write the above expression as follows.
gcd(2, 6) =
= 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. e.g.24 We want to show that there exist two integers
x and y such that gcd(6, 32) = 6x + 32y
e.g.24 i
k[i] = j[i].q[i] + r[i]
k = j.q + r
k[i] k j q r
j[i]
q[i]
r[i]
y[i]
x[i]
gcd(70, 102) = 70x + 102y
102 =
102 70 1 32 2 6 5 3 11
-16
Why is it correct?
gcd(32, 70) = 32x + 70y 1
70 = -5 11
This is correct.
gcd(6, 32) = 6x + 32y 2
32 = 1 -5
gcd(2, 6) = 2x’ + 6y’
gcd(2, 6) = 2x + 6y 1 3
6 =
gcd(102, 70) = gcd(2, 0) = 2

66. e.g.24 We want to show that there exist two integers
x and y such that gcd(6, 32) = 6x + 32y
e.g.24
gcd(6, 32) = 6x + 32y 2
32 =
gcd(2, 6) = 2x’ + 6y’
2x + 6y = gcd(2, 6) 3
6 =

67. e.g.24 We want to show that there exist two integers
x and y such that gcd(6, 32) = 6x + 32y
e.g.24
x = y’ – 5x’ and y = x’
According to x’, y’ and 5, we can
find the exact values of x and y.
Consider gcd(6, 32)
= gcd(2, 6)
= 2x’ + 6y’
= (32 – 6.5) x’ + 6y’
= 32x’ – 6.5.x’ + 6y’
This is the step we used in the Extended GCD algorithm.
= 6y’ – 6.5.x’ + 32x’
= 6(y’ – 5.x’) + 32x’
= 6x + 32y
where x = y’ – 5x’ and y = x’
Next, we want to prove this is also correct. 2
32 =
gcd(6, 32) = 6x + 32y 3
6 =
gcd(2, 6) = 2x’ + 6y’
We have already proved that this is correct.
Note that gcd(6, 32) = gcd(2, 6)

68. e.g.24 We want to show that there exist two integers
x and y such that gcd(6, 32) = 6x + 32y
e.g.24
x = y’ – 5x’ and y = x’
According to x’, y’ and 5, we can
find the exact values of x and y. 2
32 =
gcd(6, 32) = 6x + 32y 3
6 =
gcd(2, 6) = 2x’ + 6y’

69. We want to show that there exist two integers
x and y such that gcd(6, 32) = 6x + 32y
x = y’ – 5x’ and y = x’
According to x’, y’ and 5, we can
find the exact values of x and y. 2
32 =
gcd(6, 32) = 6x + 32y 3
6 =
gcd(2, 6) = 2x’ + 6y’

70. We want to show that there exist two integers
x and y such that gcd(6, 32) = 6x + 32y
x = y’ – 5x’ and y = x’
According to x’, y’ and 5, we can
find the exact values of x and y. 2
32 =
gcd(6, 32) = 6x + 32y 3
6 =
gcd(2, 6) = 2x’ + 6y’ i
k[i] = j[i].q[i] + r[i]
k = j.q + r
k[i] k j q r
j[i]
q[i]
r[i]
y[i]
x[i]
102 =
102 70 1 32 2 6 5 3 1
70 =
y = x’ 2
32 = 1 -5
0 – 5.1 y x
x = y’ – 5x’ 1 3
6 = y’ x’

71. e.g.25 (Page 48) Illustration of Theorem 2.15
Theorem 2.14 Given two integers 27, 58,
Euclid’s extended GCD algorithm computes (1) gcd (27, 58), and (2) two integers x, y such that x + 58y = gcd(27, 58)
e.g.25 (Page 48)
Illustration of Theorem 2.15
Theorem 2.15 Two 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 x + 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)
By Theorem 2.14, we know that there are integers x, y such that x + 58y = 1

72. e.g.25 Illustration of Theorem 2.15
Lemma 2.11
If there exist integers x, y such that 27x + 58y = 1,
then gcd(27, 58) = 1
(i.e., 27 and 58 are relatively prime.)
Illustration of Theorem 2.15
Theorem 2.15 Two 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 x + 58y =1
Why is it correct? If
We know that there are integers x, y such that x + 58y = 1
By Lemma 2.11, we know that gcd(27, 58) = 1

73. e.g.26 (Page 49)
Corollary 2.16 Consider 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 .7 x = 1 has a solution in Z7
if and only if
there exist integers x, y such that 2x + 7y = 1
Theorem 2.15 Two positive integers 2, 7 have gcd(2, 7) = 1 (and thus they are relatively prime) if and only if there are integers x, y such that x + 7y =1

74. e.g.26
Corollary 2.16 Consider 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 .7 x = 1 has a solution in Z7
if and only if
there exist integers x, y such that 2x + 7y = 1
2 has a multiplicative inverse in Z7
Theorem 2.15 Two positive integers 2, 7 have gcd(2, 7) = 1 (and thus they are relatively prime) if and only if there are integers x, y such that x + 7y =1
gcd(2, 7) = 1

75. Multiplicative inverse1 2 3 4
Multiplicative inverse
e.g.26
Since gcd(3, 5) = 1,
3 has the multiplicative inverse in Z5
Z5:

76. e.g.26 a 1 2 3 4 Multiplicative inverse Z5: a 1 2 3 4 5X
Since gcd(3, 6) = 2  1,
3 has no multiplicative inverse in Z6
Z6:

77. e.g.26 a 1 2 3 4 Multiplicative inverse Z5: a 1 2 3 4 5X
Z6: a 1 2 3 4 5 6
Multiplicative inverse
Z7: a 1 2 3 4 5 6 7
Multiplicative inverse X
Z8: a 1 2 3 4 5 6 7 8
Multiplicative inverse X
Z9:

78. e.g.27 (Page 49)
Corollary 2.17 Note 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
We know the following corollary.
Corollary 2.16 Consider 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.

79. e.g.27 a 1 2 3 4 Multiplicative inverse Since 5 is a prime number,
every non-zero a  Z5 has a multiplicative inverse.
Z5: a 1 2 3 4 5
Multiplicative inverse X
Z6: a 1 2 3 4 5 6
Multiplicative inverse
Since 5 is a prime number,
every non-zero a  Z5 has a multiplicative inverse.
Z7: a 1 2 3 4 5 6 7
Multiplicative inverse X
Z8: a 1 2 3 4 5 6 7 8
Multiplicative inverse X
Z9:

80. e.g.27 (Page 52) Illustration of Corollary 2.18
If there exist integers x, y such that 2x + 7y = 1,
then the multiplicative inverse of 2 in Z7 is
x mod 7
Illustration of Corollary 2.18
Corollary 2.18 If 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 = 1 The inverse of 2 in Z7 is equal to
x mod 7
Why is it correct?

81. x = -3
y = 1
The algorithm finds x +7y = 1 (i.e., 2(-3) + 7(1) = 1)
e.g.28 (Page 52)
The multiplicative inverse of 2 in Z7 is mod 7 = 4
We want to find the multiplicative inverse of 2 in Z7
Consider two integers 2 and 7
k = j.q + r i
k[i] = j[i].q[i] + r[i] k j q r
k[i]
j[i]
q[i]
r[i]
y[i]
x[i]
0-3.1
7 = 7 2 3 1 1 -3 1
2 = 2 1 2 1
gcd(2, 7) = gcd(1, 0) = 1
This implies that there exists a multiplicative inverse of 2 in Z7