Download Chapter 3: Primes and their Distribution

Chapter 3: Primes and their Distribution
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Chapter 3: Primes and their Distribution
SECTION B Testing Numbers
By the end of this section you will be able to
 understand
B1 Floor and Ceiling Function
The floor function is defined as:
Definition (3.7).
The floor function is denoted by  x  and is the greatest integer less than or equal to x.
The domain and range of the floor function are
and
respectively. We have
 x  : 
For example
7.1  7 ,    3 , e  2 , 2.1  2 , 2.1  3
The graph of the floor function is
Fig 2
Next we define the ceiling function.
Definition (3.8).
The ceiling function is denoted by  x  and is the least integer greater than or equal to x.
The domain and range of the floor function are
and
respectively. We have
 x  :  .
For example
7.1  8 ,    4 , e  3 , 2.1  3 , 2.1  2
The graph of the floor function is
Chapter 3: Primes and their Distribution
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B2 Testing of Composite Numbers
How can we test whether a given number is prime or composite?
The following proposition is a useful test for small composite numbers:
Proposition (3.9).
If n is composite then it has a divisor d such that 1  d   n  .
How do we prove this result?
By contradiction.
Proof.
We are given that n is composite. This means there exists integers 1  d1  n and 1  d2  n
such that
n  d1d 2
Suppose d1  n and d2  n . We have
n  d1d2  n n  n
This is impossible. Our supposition must be wrong so one of d1 or d 2 must be  n .
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Examples are
315  32  5 7 
983  983 (a prime number)
1001  7 1113
Note that for testing whether 983 is prime we need to see if all the numbers from 2 to
  983   31 go into 983. This is still a tedious task.
Can we simplify the above test?
Yes.
Chapter 3: Primes and their Distribution
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Corollary (3.10).
If n is composite then it has a prime divisor p such that p   n  .
Proof.
By the above Proposition (3.9):
If n is composite then it has a divisor d such that 1  d   n  .
we have that n has a divisor d such that 1  d   n  . By the Fundamental Theorem of
Arithmetic there exists a prime p such that p d . Hence p   n  .
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How does this simplify the test for composite numbers?
Well for testing whether 983 is prime or composite we only need to see if the primes less
than or equal to  983   31 go into 983. The primes  31 are
2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31
There are 11 primes  31 and we only need to see if these are proper divisors of 31. By the
earlier proposition (3.9) we would have had to check whether all the numbers between 2
and 31 are divisors of 983.
Corollary (3.10) means that to test a given positive integer n for primality we only have to
divide by the prime numbers between 2 and  n  .
Example 1
Test 1001 for prime. If 1001 is composite write down its prime decomposition
Solution
We first find  1001  31.63858404   31 . We only need to check the primes:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31
2 and 3 clearly do not go into 1001 exactly. What about 7?
Well 7 143  1001. This means that 1001 is composite. What are the other divisors of
1001?
To check if 143 is prime we only need to if the primes less than or equal to  143   11
are divisors of 143.
7 is not a divisor of 143. What about 11?
1113  143
Hence the prime divisors of 1001 are 7, 11 and 13:
1001  7 1113
B3 Eratosthenes Sieve
The diagram below shows the Sieve of Eratosthenes for integers less than or equal to 100.
First we write down all the integers between 2 and 100. Since 2 is prime we cross out all
the multiples of 2 or the even numbers apart from 2 itself. The first of the remaining
integers is 3 so it must be a prime. Now we cross out all the multiples of 3 apart from 3
itself. We continue this process for 5, 7, 11, 13, … The integers which do not fall into this
sieve are the prime numbers:
Chapter 3: Primes and their Distribution
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