Primes. The Fundamental Theorem of Arithmetic.
CSI30
Prime Number - is a positive integer greater than 1 that is divisible only by
1 and itself.
primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, ...
Composite Number – is a positive integer greater than 1, that is divisible
by at least one more integer number other than 1 and itself.
composites: 4, 6, 8, 9, 10, 12, 14, 15, ....
In what sense are the primes the building blocks of positive integers?
Prime Factorization of n A product of primes written that equals n, written in increasing order.
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Primes. The Fundamental Theorem of Arithmetic.
CSI30
In what sense are the primes the building blocks of positive integers?
[Theorem 1] The Fundamental Theorem of Arithmetic
Every positive integer greater than 1 has a unique prime factorization.
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Primes. The Fundamental Theorem of Arithmetic.
CSI30
How do you check if a number is prime?
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Primes. The Fundamental Theorem of Arithmetic.
CSI30
[Theorem 2]
If n is a composite integer, then n has a prime divisor less than or equal to  n.
Example 3: Show that 103 is a prime number.
Solution:
 103≈10.15
Therefore the only possible divisors are 2, 3, 5, 7.
103 is not divisible by any of them. Thus 103 is a prime number.
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Primes. The Fundamental Theorem of Arithmetic.
CSI30
[Theorem 2]
If n is a composite integer, then n has a prime divisor less than or equal to
n .
Proof:
If n is composite, then (by the def. of composite number) n has a factor a,
with 1< a < n. Therefore n = a  b, where b  Z+.
Let's show that a   n or b   n :
If both a >  n and b >  n , then ab >  n  n = n – which is a contradiction.
Therefore, a   n or b   n and since a and b are divisors of n (they are
its factors) we see that n has a positive divisor   n .
By the Fundamental Theorem of Arithmetic, this positive divisor is either
prime or has a prime divisor less than itself. In either case, n has a prime
divisor   n .
qed
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Primes. The Fundamental Theorem of Arithmetic.
CSI30
Extra! ….
[Theorem 3]
There are infinitely many primes
Proof: (by contradiction)
Let's assume that there is a finite number of primes, p1, p2, ..., pn
Let q = p1 p2 ... pn + 1
By the Fundamental Theorem of Arithmetic, q is either a prime number or can be written as a
product of two or more primes.
None of the primes divide it, since if there is a prime pi that divides it (pi|q), then pi should
divide (q - p1 p2 ... pn) , but q - p1 p2 ... pn = 1 and pi | 1.
Thus, q is a prime number, and it is not from our list of n primes – contradiction.
Our assumption was wrong, there are infinitely many primes.
qed
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Greatest Common Divisor (GCD) and
Least Common Multiples (LCM)
CSI30
[Def]
Let a, b  Z, a,b  0. The largest d  Z such that d | a and d | b is called
the greatest common divisor of a and b
denotation: gcd(a,b)
How to find GCD
1. find prime factorization of both a, and b
2. take the prime factors present in both factorization with their smallest
powers (present in those factorization)
3. multiply
Example 6:
Find gcd(28,72).
Solution:
28 = 22  7, 72 = 23  32
gcd(28,36) = 22 = 4.
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Greatest Common Divisor (GCD) and
Least Common Multiples (LCM)
CSI30
Example 7:
Find gcd(330,420).
Solution:
330 = 2  3  5  11,
420 = 22  3 5  7
gcd(330,420) = 2  3  5 = 30.
[Def]
Two integers are relatively prime if their GCD is 1.
Integers a1, a2, ..., an are pairwise relatively prime if gcd(ai,aj) = 1 for all
pairs of i,j such that 1 i < j  n
Example 8: Determine if each of these sets are pairwise relatively prime
a) 11, 15, 19
b) 14, 15, 21
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The Euclidean Algorithm
CSI30
The method describes an algoirthm for finding the GCD (using prime
factorizations) is quite inefficient.
Euclidean Algorithm gives a more efficient way of finding GCD.
It is named after Greek mathematician Euclid.
procedure gcd(a, b: positive integers)
x := a
y := b
while y  0:
r := x mod y
x := y
y := r
{gcd(a,b) is x}
This algorithm uses the following lemma:
[Lemma 1]
Let a=bq+r, where a,b,q, and r are integers. Then gcd(a,b) = gcd(b,r)
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The Euclidean Algorithm
CSI30
Example 4: find gcd(1001,1331)
Solution: gcd(1001,1331) = gcd(1331,1001)
– we prefer to take smaller as divisor and larger as dividend.
1331
=1 R 330
1001
Thus gcd(1331,1001) = gcd(1001,330) = gcd(330,11) = gcd(11,0) STOP
330
1001
=30 R 0
=3 R 11
11
330
Answer: gcd(1001,1331) = 11
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