MATH 57091 - Algebra for High School Teachers
Cancellation & Multiplicative Inverses
Professor Donald L. White
Department of Mathematical Sciences
Kent State University
D.L. White (Kent State University)
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Cancellation
Recall that when we discussed algebraic properties of congruences,
we did not include dividing both sides of a congruence by an integer,
or “cancellation” of an integer factor from both sides.
There are very good reasons for this.
Example: Observe that 20 ≡ 6 (mod 14); that is,
2(10) ≡ 2(3) (mod 14).
However, 14 - 10 − 3, and so
10 6≡ 3 (mod 14).
From the point of view of divisibility, 14 | 20 − 6, so 14 | 2(10 − 3).
But we have seen that n | ab does not imply n | a or n | b.
Euclid’s Lemma says if n | ab and (n, a) = 1, then n | b.
In this example, (14, 2) 6= 1 and in fact 14 - 10 − 3.
We use Euclid’s Lemma to derive a “cancellation” theorem.
D.L. White (Kent State University)
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Cancellation
Theorem (Cancellation)
Fix a positive integer n and let a, b, and c be integers.
If ab ≡ ac (mod n) and (a, n) = 1, then b ≡ c (mod n).
Proof: Since ab ≡ ac (mod n), we have n | ab − ac; that is, n | a(b − c).
If (a, n) = 1, then Euclid’s Lemma implies n | b − c,
and therefore b ≡ c (mod n).
The theorem says we can “cancel” any integer factor from both sides,
as long as it is relatively prime to the modulus.
But what we really want to do is “divide” both sides by the same integer.
D.L. White (Kent State University)
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Multiplicative Inverses
Let’s look at how we solve a linear equation by “cancellation.”
For example, solve 3x = 21 for x.
We could say that 3x = 3 · 7 and “cancel” the 3 from both sides.
Again, we are really dividing both sides of the equation by 3:
3x
21
= .
3
3
But division does not make sense in some situations.
For example, how would we solve the matrix equation AX = B, where
2 5
3 4
and B =
?
A=
1 3
1 2
Since matrix multiplication is not commutative,
“division” of matrices is not possible. Instead, we multiply both sides
of the equation by the multiplicative inverse A−1 of A:
3 −5
3 4
4 2
X = A−1 B =
=
.
−1
2
1 2
−1 0
D.L. White (Kent State University)
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Multiplicative Inverses
Since we cannot really “divide” with congruences,
we need to think in terms of multiplying by multiplicative inverses.
Generally, a multiplicative inverse is something we multiply by
to get the “multiplicative identity.” In this spirit, we define
Definition
Fix a positive integer n and let a be an integer.
We say that an integer b is a multiplicative inverse of a modulo n if
ba ≡ 1 (mod n).
NOTE: A multiplicative inverse b for a is not unique. In particular,
if ba ≡ 1 (mod n) and c ≡ b (mod n), then ca ≡ 1 (mod n) also.
D.L. White (Kent State University)
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Existence of Inverses
Which integers a have a multiplicative inverse mod n?
a has an inverse ⇐⇒ ba ≡ 1 (mod n) for some integer b
⇐⇒ n | ba − 1 for some integer b
⇐⇒ ba − 1 = cn for some integers b, c
⇐⇒ ba + (−c)n = 1 for some integers b, −c
⇐⇒ (a, n) = 1.
Therefore, we have the following theorem.
Theorem
Fix a positive integer n.
An integer a has a multiplicative inverse mod n if and only if (a, n) = 1.
D.L. White (Kent State University)
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Finding an Inverse
The next question is how to find a multiplicative inverse for a mod n.
For small values of n, an inverse can often be found “by inspection.”
Note that if b is an inverse for r and a ≡ r (mod n),
then b is also an inverse for a, and a, r are both inverses for b.
Hence it is necessary only to find inverses for integers
r , 0 6 r < n, with (r , n) = 1.
Moreover, each r with 0 6 r < n and (r , n) = 1
has exactly one inverse r 0 mod n with 0 6 r 0 < n.
D.L. White (Kent State University)
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Finding an Inverse
Example: Find the inverses mod 14 for all invertible integers.
Every integer that is invertible mod 14 is congruent mod 14 to one of
1, 3, 5, 9, 11, or 13.
It is clear that 1 is the inverse of 1 mod n for any n.
The inverse of 3 must be one of 3, 5, 9, 11, or 13.
Which of these satisfies the congruence 3 ·
≡ 1 (mod 14)?
Since 3 · 5 = 15 ≡ 1 (mod 14), 3 and 5 are inverses of each other mod 14.
The inverse of each of 9, 11, 13 is one of the integers 9, 11, 13.
Check that 9 and 11 are inverses of each other,
which forces 13 to be its own inverse.
D.L. White (Kent State University)
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Finding an Inverse
For large values of n, finding inverses by inspection is not practical.
Recall that a has an inverse b mod n if and only if
ba + cn = 1 for some integer c.
Hence we can use the Euclidean Algorithm to show (a, n) = 1,
and then use those equations to write 1 as a combination of a and n.
Example: Find a multiplicative inverse for 275 mod 562.
562 = 2(275) + 12
(1)
275 = 22(12) + 11
(2)
12 = 1(11) + 1
(3)
11 = 1(11)
(4)
1 = 12 − 1(11)
= 12 − 1[275 − 22(12)] = 23(12) − 1(275)
= 23[562 − 2(275)] − 1(275) = (−47)(275) + 23(562)
Hence −47, or 562 − 47 = 515, is an inverse for 275 mod 562.
D.L. White (Kent State University)
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