Greatest Common Divisors
and Prime Factors!
CSCI 2824, Fall 2011!
http://l3d.cs.colorado.edu/~ctg/
classes/struct11/Home.html
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Assignments
• To read this week: Sections 2.5-2.7 (Emsley/
Crawley)
• For next week: more number theory (no assigned
reading!)
• Problem Set 1 due Tuesday (hard copy please!!!) !
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Divisors
The set of divisors of a positive integer n are all
those integers d whose absolute value is less than
or equal to n and that have the property: d|n
So, for example, the divisors of 18 are:
+/- 1, +/- 2, +/- 3, +/- 6, +/- 9, +/- 18
The divisors of 42 are:
+/- 1, +/- 2, +/- 3, +/- 6, +/- 7, +/- 14, +/- 21, +/42 Primes and GCDs
A prime number is a positive integer >= 2
whose only divisors are 1 and itself.
For any two numbers m and n, gcd(m, n) is the
greatest common divisor shared by m and n.
gcd(14, 35) = 7
gcd(6, 0) = 6 gcd(8, 9) = 1
gcd(7, 5) = 1
Relatively Prime Numbers
Two numbers are relatively prime if their gcd
is 1:
8 and 9 are relatively prime.
12 and 15 are not relatively prime.
Any two distinct primes (e.g., 5 and 11) are
relatively prime.
1 and any number are relatively prime.
A couple of questions
What numbers can be expressed in the form:
12 x + 27 y
where x and y are integers?
A couple of questions
What numbers can be expressed in the form:
13 x + 27 y
where x and y are integers?
A hypothesis
If two numbers m and n are relatively prime,
then we can find integers x and y such that:
m*x + n*y = 1
This implies in turn that any integer can be
expressed in combinations of m and n.
Bezout s Theorem
If two numbers m and n are relatively prime, then we
can find integers x and y such that:
m*x + n*y = 1
Proof by contradiction:
• Suppose that d is the smallest positive integer such
that m*x + n*y = d, and suppose that d > 1. We ll
find a positive integer smaller than d that can be
expressed as a combination of m and n.
Okay, let s get started…
m * x + n * y = d where d > 1
• Now, d can t divide both m and n, since then they
wouldn t be relatively prime.
• So let s say d does not divide m. Then we can
write:
m = d*q + r where 0 < r < d
• So now we can write:
r
=
m – d*q
=
m – (m*x + n*y)*q
=
m(1 – q*x) - n * (q*y)
Okay, let s get started…
m * x + n * y = d where d > 1
• So now we can write:
r
=
m – d*q
=
m – (m*x + n*y)*q
=
m(1 – q*x) - n * (q*y)
• But hold it! We just showed that r is a combination
of m and n. And we know that r is less than d! So
our original premise can t be right.
Bezout s Theorem, Again
If two numbers m and n are relatively prime,
then we can find x and y such that:
m*x + n*y = 1
Try it for any two relatively prime numbers:
7 and 26
9 and 11
A Corollary to Bezout s
Theorem
If gcd(m, n) = d where d > 1, then we can find
an integer combination of m and n that sums
to d:
m * x + n * y = d
Why? Note that gcd(m/d, n/d) = 1, and then
we can find a combination such that:
(m/d) * x + (n/d) * y = 1
How to find the gcd? (Euclid again!)
We want to find gcd(a, b) where a >= b.
First, we note that if a = b, the answer is a.
Also, if b = 0, the answer is a.
Otherwise, we can find gcd(a, b) by noting
that the greatest common divisor of a and b
is also the greatest common divisor of b and
the remainder r when a is divided by b. Why does this work?
Suppose d|a and d|b. (In prose, d goes into both a and b.)
And suppose a = b*q + r.
Then d|b and d|(a – b*q).
Two examples:
gcd (99, 18) =
gcd(18, 9) =
gcd(9, 0) = 9
gcd(70, 49) =
gcd(49, 21) =
gcd(21, 7) =
gcd(7, 0) = 7 Scheme version of the algorithm
for finding the GCD
!
(define (gcd a b)!
(if !(= b 0) !
! !
!a!
! !
!(gcd b (remainder a
b))))!
!
Not only is this idea super-fast, but it s
also useful:
(18, 7) = (7, 4) = (4, 3) = (3, 1) = (1, 0) = 1
18 = (2 * 7) + 4
7 = (1* 4) + 3
4 = (1 * 3) + 1
Now, unwind those statements:
1 = 4 – (1 * 3)
= 4 - (1 * (7 – (1 * 4))) = 4 – (7 – 4) = (2 * 4) - 7
= ((2 * (18 – (2 * 7))) – 7 = (2 * 18) – (5 * 7)
The Structure of Divisors
Let s try to break down a number s divisors
as far as we can:
84 = 2 * 42
= 2 * 2 * 21
= 2 * 2 * 3 * 7
We can t go any further than this, since we
have only prime factors.
The Fundamental Theorem of
Arithmetic
Every number n has a unique
factorization into primes.
A Lemma
Suppose d|(ab) and gcd(d,a) = 1
In prose: d goes into the product of a and
b, but d and a are relatively prime.
Then d|b
Let s prove the lemma.
Since
d|ab
We can write:
ab = dq
And since gcd(d, a) = 1
We can find x and y such that:
dx + ay = 1
So: dbx + aby = b
dbx + dqy = b
d(bx + qy) = b
Lemma 2 (a corollary to Lemma 1)
We now know that if d|(ab) and gcd(d,a) = 1,
then d|b. Suppose d is a prime number (let s
call it p for prime). Then if p|(ab), we know
that either p|a or p|b.
Proof: Suppose p doesn t go into a. Then by
the definition of greatest common divisor
(and the definition of a prime number ),
gcd(p,a) = 1. Which means, from our
previous lemma, that p|b.
More generally, we have the following:
p|q1q2q3…qn
means that p must go into at least one factor in
the product. Now we re ready to prove the
Fundamental Theorem of
Arithmetic
First we show that every number can be factored into primes.
(We ll leave uniqueness for the next step.)
Step 1. We know that 2 can be factored into primes (i.e., 2
itself).
Step 2. Suppose that all numbers from 2…n can be factored
into primes. Consider n+1. If it s prime, we re done.
Step 3. Suppose it isn t prime. Then it must have some factor
q. So qx = (n+1). But we know that both q and x are
smaller than n+1, so they can be factored into primes.
A note: this technique (step 2) is called strong induction.
The factorization is unique (proof by
contradiction).
Suppose the factorization weren t unique. Let m be the
smallest number with two distinct prime factorizations,
with factors ordered low to high:
p1p2p3..pj =
m
=
q1q2q3…qk
If p is equal to q, then we can divide both factorizations by
that number, and we have a smaller value with two distinct
factorizations (contrary to our assumption). So choose the
smaller of p, q: let s say it s p. Then by our earlier lemma
2, p must go into one of the q factors, contrary to our
assumption that they re all prime! Either way, we have a
contradiction. Okay, now we re ready to play
with factorizations:
n = 2e1 * 3e2 * 5e3 * 7e4 *…