Discrete Mathematics with Applications
MATH236
Dr. Hung P. Tong-Viet
School of Mathematics, Statistics and Computer Science
University of KwaZulu-Natal
Pietermaritzburg Campus
Semester 1, 2013
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Table of contents
1
Extended Division Algorithm
2
Multiplicative Inverses in Zm
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Extended Division Algorithm
Extended Division Algorithm
Lemma
Let S be a nonempty set of integers that is closed under addition and
subtraction. The exactly one of the following statements is true:
S = {0}
S = {0, ±d, ±2d, ±3d, · · · }, where d is the smallest positive integer
in S.
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Extended Division Algorithm
Extended Division Algorithm
Theorem
Let a and b be nonzero integers. Then gcd(a, b) can be expressed as a
linear combination of a and b with integers coefficient, that is,
gcd(a, b) = sa + tb
for some integer s, t.
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Extended Division Algorithm
Extended Division Algorithm
Example
Express gcd(112, 268) as an integer linear combination of 112 and 268.
We have that gcd(112, 268) = 4 = r3
Now 4 = 24 − 20
Replace r2 = 20 by 44 − 24
Then 4 = 24 − (44 − 24) = 2 · 24 − 44
Replace r1 = 24 by 112 − 2 · 44
Then 4 = 2 · 112 − 5 · 44
Replace r0 = 44 by 268 − 2 · 112
We obtain 4 = 2 · 112 − 5(268 − 2 · 112) = 12 · 112 − 5 · 268.
Consequently, 4 = 12 · 112 + (−5) · 268
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Multiplicative Inverses in Zm
Multiplicative inverses in Zm
Let m ≥ 1 be an integer
Let Zm = {0, 1, 2, , · · · , m − 1} be the set of all remainders when
divided by m
For a ∈ Zm , if there exists b ∈ Zm such that
ab = ba ≡ 1(mod m),
then b is said to be a multiplicative inverse of a in Zm and is denoted
by a−1 .
Given a ∈ Zm , it is not certain that the multiplicative inverse of a
exists
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Multiplicative Inverses in Zm
Multiplicative inverses in Zm
Example
Consider 2 ∈ Z4 = {0, 1, 2, 3}
2×0 = 0
2×1 = 2
We have
2×2 = 0
2×3 = 2
Thus 2 has no multiplicative inverse in Z4
However, in Z5 , we have 2 × 3 = 3 × 2 = 6 ≡ 1(mod 5)
Thus 2−1 = 3 in Z5
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Multiplicative Inverses in Zm
Existence of multiplicative inverses in Zm
Theorem
A number a ∈ Zm has a multiplicative inverse in Zm if and only if
gcd(a, m) = 1.
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Multiplicative Inverses in Zm
Existence of multiplicative inverses in Zm
Proof.
Suppose first that a and m are relatively prime
By the Extended Division Algorithm, there exists integers s, t such
that sa + tm = 1
Write s = qm + r with 0 ≤ r < m. Then r ∈ Zm
We have ar = a(s − qm) = as − aqm = 1 − tm − aqm =
1 − m(t + aq) ≡ 1(mod m)
Thus r is the multiplicative inverse of a in Zm
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Multiplicative Inverses in Zm
Existence of multiplicative inverses in Zm
Proof.
Conversely, suppose first that a has a multiplicative inverse b in Zm
and there exist t, a0 , m0 ∈ {2, 3, · · · , m − 2} such that a = ta0 and
m = tm0
Since ab ≡ 1 (mod m), a0 tb ≡ 1 (mod m)
Multiplying both sides by m0 , we obtain a0 btm0 ≡ m0 (mod m)
Since m = tm0 , we have that a0 bm ≡ m0 (mod m)
Clearly a0 bm ≡ 0 (mod m) so m0 ≡ 0 (mod m), which is impossible
Hence, gcd(a, m) = 1 as wanted.
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Multiplicative Inverses in Zm
Compute the multiplicative inverses in Zm
Example
Find 4−1 in Z21 .
We have gcd(4, 21) = 1 so 4−1 exists in Z21 .
As 21 = 4 · 5 + 1, we obtain that 21 − 4 · 5 − 1
Hence, 4−1 = −5 = 21 − 5 = 16
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Multiplicative Inverses in Zm
Compute the multiplicative inverses in Zm
Example
Find 13−1 in Z73 .
We first find gcd(13, 73). We have
73 = 13 · 5 + 8
13 = 8 + 5
8 = 5+3
5 = 3+2
3 = 2+1
Thus gcd(13, 73) = 1. Hence 13−1 exists in Z73 .
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Multiplicative Inverses in Zm
Compute the multiplicative inverses in Zm
We have
1 = 3−2
= 3 − (5 − 3)
= 3·2−5
= 2 · (8 − 5) − 5
= 2·8−3·5
= 2 · 8 − 3 · (13 − 8)
= 5 · 8 − 3 · 13
= 5 · (73 − 13 · 5) − 3 · 13
= 5 · 73 − 28 · 13
Hence 13−1 = 73 − 28 = 45 in Z73 .
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Multiplicative Inverses in Zm
Uniqueness of multiplicative inverses in Zm
Theorem
If a has a multiplicative inverse in Zm , then this inverse is unique, i.e., if
x, y ∈ Zm such that ax ≡ 1 (mod m) and ay ≡ 1 (mod m), then x = y .
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Multiplicative Inverses in Zm
Uniqueness of multiplicative inverses in Zm
Proof.
Suppose that x, y ∈ Zm such that
ax ≡ 1 (mod m) and ay ≡ 1 (mod m)
Then there exist s, t ∈ Z such that ax = sm + 1 and ay = tm + 1
Hence, a(x − y ) = m(s − t)
By assumption, a has at least one inverse a−1 , so
x − y = a−1 (s − t)m and thus
x − y ≡ 0 (mod m)
Since x, y ∈ Zm , the only possibility is that x = y .
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