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2. Prime Numbers
You probably know that the numbers 2, 3, 5, 7, 11, 13 and 17 are prime numbers. We give a precise definition of the notion of prime number below.
Definition 2.1: A number p ∈ ` is a prime number if and only if p ≥ 2 and
the only natural numbers which divide p are 1 and p.
The prime numbers between 1 and 100 are given by 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97.
Prime numbers and their properties were first studied by the ancient Greek mathematicians of Pythagoras's school from 500 BC to 300 BC. Book IX of Euclid’s Elements (300 BC) contains a proof that there are infinitely many prime numbers. Euclid also proved the Fundamental Theorem of Arithmetic; namely that every natural number can be written in a unique way as a product of primes. There are a multitude of mathematical results related to prime numbers. Interestingly though, there is no formula for easily determining whether a natural number is a prime number, and it is very difficult to determine (even with a high powered computer) whether a very large natural number is prime. The largest known prime was found by the Great Internet Mersenne Prime Search in November 2003 (and perhaps not the most recent by the time you read these notes) has 6,320,430 decimal digits. That’s HUGE! In fact, it could take over 1,500 pages to write this prime number in a 12 point font!
Definition 2.2: Let n ≥ 2 be a natural number. A prime factorization of n has the form 1 2 1 2 k n p p pk = α^ α"^ α
where p 1 , p 2 ,..., pk are prime numbers and α 1 , α 2 ,..., α ∈ k `.
Theorem 2.3: Every natural number n ≥ 2 has a unique prime factorization.
Proof: If n is prime, there is nothing to prove. So, suppose n is not prime. Then, there exists an integer k such that 1 < k < n and k divides n. Pick the smallest such integer and call it p 1. If p 1 were not prime, then it would have a divisor q
such that 1 < q < p 1. But, q divides p 1 implies that q divides n , which contradicts
to the minimality of p 1. Thus, p 1 is prime. Write n = p m 1 1. If m 1 is prime, we
have the desired representation. Otherwise, with the same argument, we obtain p 2 such that m 1 (^) = p m 2 2 , and this implies that n = p p m 1 2 2.
The decreasing sequence n > m 1 (^) > m 2 > ... > 1 can not continue indefinitely, so at
some point mk (^) − 1 is a prime; call it pk. Then n = p p 1 2 (^) ... pk is the prime
factorization of n.
To show uniqueness, assume that we have two prime factorizations. n = p p 1 2 (^) ... pr = q q 1 2 ... qs
We can assume that r ≤ s , and that our primes are written in an increasing way; p 1 (^) ≤ p 2 (^) ≤ ... ≤ pr and q 1 (^) ≤ q 2 ≤ ...≤ qs.
Since p 1 divides q q 1 2 (^) ... q (^) s , and the qi are prime, we must have p 1 (^) = qk for some
k , but then p 1 (^) ≥ q 1. The same reasoning gives q 1 (^) ≥ p 1 , so p 1 (^) = q 1. Cancel this
common factor to get p p 2 3 (^) ... pr = q q 2 3 ... qs. Continuing in this way, we can
divide by all pi and get 1 = qr (^) + 1 qr (^) + 2 ... qs. But the qi were assumed to be greater
than or equal to 1, so they must be equal to 1. Then r = s and pi = qi for all i ,
making two factorizations equal.
The following result can help determine whether a natural number is prime, and lessen the pain of finding a prime factorization.
Theorem 2.4: Let n ≥ 2 be a natural number and suppose k ∈ ` so that
k^2 > n. The number n is a prime number if and only if no prime number less than k divides n.
Proof: ( ⇒) Suppose n is prime. Then by definition of prime numbers, its only
divisors are 1 and n. Thus, n is not divisible by any of the prime numbers less
than k (note that n itself is not one of those numbers since k^2 > n ).
( ⇐) Suppose none of the prime numbers less than k divides n.
Then, any natural number(except 1 of course!!!) less than or equal to k does not divide n (since any such natural number is either a prime number less than k or a product of prime numbers which are less than k ) Let p be a natural number which is bigger than k ( p ≠ n ). Suppose p divides n.
There exists a natural number q such that n = q ⋅ p. Since k^2 > n , we have;
q ⋅ p < k^2. We know that p > k , so q must be smaller than k. Hence, q is a
natural number that divides n and it is smaller than k. But this is a contradiction to our hypotheses since q is either a prime number less than k or is a product of prime numbers which are less than k. Hence, n does not have any divisors other than 1 and itself; n is prime.
Example 2.5: Show that the numbers 2, 3, 5 and 7 can be used to determine all of the prime numbers smaller than 100.
Solution: Let n ∈ ` with n < 100. Notice that 10 2 = 100 > n , and the prime numbers less than 10 are 2, 3, 5 and 7. So, Theorem 2.4 tells us that n is a prime number if and only if the numbers 2, 3, 5 and 7 do not divide n.
Remark 2.6: The example above can be used to identify the prime numbers
Example 2.9: Give the prime factorization of 2303028.
Solution: 2303028 is a multiple of 4 since its last two digits are divisible by 4. Also,
( ) ( ) ( )
2 2303028 = 4 575757 = 2 575757
Notice that the digits of 575757 sum to 36, and 36 is a multiple of 9. As a result, 575757 is a multiple of 9. In fact,
( ) ( ) ( )
2 575757 = 9 63973 = 3 63973
So,
( ) ( ) ( ) ( ) ( ) ( )
2 2 2 2303028 = 4 575757 = 2 575757 = 2 3 63973
We can see that 2, 3 and 5 do not divide 63973. However, the next prime, 7,
divides this number, and 63973 = 7 9139( ). This gives
( ) ( ) ( )( )
2 2 2303028 = 2 3 7 9139
Neither 7 nor 11 divide 9139, but 13 divides this number, and 9139 = 13 703( ).
Combining this information gives
( ) ( ) ( )( )( )
2 2 2303028 = 2 3 7 13 703
13 and 17 do not divide 703, but 19 divides this number, and 703 = 19 37( ).
Since 37 is prime, we have the prime factorization
( ) ( ) ( )( )( )( )
2 2 2303028 = 2 3 7 13 19 37
Remark 2.10: The example above illustrates that it can be very painful to find the prime factorization of a large natural number. It can even be difficult for computers to find the prime factorization of extremely large natural numbers.
Theorem 2.11: There are infinitely many prime numbers.
Proof: Assume for contradiction that there are a finite number of prime numbers. Let’s call them p 1 (^) , p 2 (^) , p 3 ,..., pN.
Consider the number p = ( p 1 (^) ⋅ p 2 (^) ⋅ p 3 ⋅ ... ⋅ pN ) + 1.
Every prime number, when divided into this number, leaves a remainder of 1. Thus, this number has no prime factors (and by our assumption it is not prime itself). This is a contradiction. Hence, there must be infinitely many prime numbers.
Exercises – Use Hand Calculations On The Exercises Below.
- Use the exercises at the end of the previous section and the list of primes at the beginning of this section to give the prime factorization of
- Use the exercises at the end of the previous section and the list of primes at the beginning of this section to give the prime factorization of
- Use Theorem 2.4 to show that 359 is prime. (Hint: 192 > 359 )
- Use Theorem 2.4 to show that 887 is prime. (Hint: 302 > 887 )
- Create a table of natural numbers between 2 and 200. Explain how the multiples of 2, 3, 5, 7, 11 and 13 (other than 2, 3, 5, 7, 11 and 13) can be used to determine the prime numbers in this table. Use your process to determine the prime numbers less than 200.
- Two natural numbers are said to be twin primes if they are both prime numbers and they differ by 2. For example, 5 and 7 are twin primes. 17 and 19 are twin primes. Use the table created in the exercise above to list all of the twin primes less than 200.
Solutions:
- 42042 is an even number, so it is divisible by 2. The sum of its digits is 12; it is divisible by 3, but not by 9; 42042 = 2 3 7007⋅ ⋅. Obviously, 7007 is divisible by 7; 42042 = 2 3 7 1001⋅ ⋅ ⋅. 1001 is not divisible by 5 since its ones digit is 1, but it is divisible by 7, 42042 = 2 3 7 7 143⋅ ⋅ ⋅ ⋅. 143 is not divisible by 7, it is divisible by 11; 42042 = 2 3 7 7 11 13⋅ ⋅ ⋅ ⋅ ⋅. Since 13 is prime, we stop here. Thus; 42042 = 2 3 7⋅ ⋅ 2 ⋅11 13 ⋅.
- 109460 is even, so it is divisible by 2. The sum of the digits is 20, so it is not divisible by 3. The last two digits form the number “60” which is divisible by 4, hence our number is divisible by 4.The ones digit is 0, so it is divisible by 5. The last three digits form “460” which is not divisible by 8, our number is not divisible by 8; 109460 = 4 5 5473⋅ ⋅. 5473 is not divisible by 7 or 11, but it is divisible by 13; 109460 = 4 5 13 421⋅ ⋅ ⋅. As we showed in Example 2.7, 421 is prime. Hence,
109460 = 4 5 13 421⋅ ⋅ ⋅.
The multiples of 2 (other than 2) are shaded red, the remaining multiples of 3 (other than 3) are shaded blue, the remaining multiples of 5 (other than 5) are shaded yellow, the remaining multiples of 7 (other than 7) are shaded green, the remaining multiples of 11 (other than 11) are shaded orange, and the remaining multiples of 13 (other than 13) are shaded pink. The numbers whose cells are not shaded are prime numbers!
- The table in Exercise 5 tells us that the twin prime numbers less than 200 are:
5 & 7, 11&13, 17 &19, 29 & 31, 41& 43, 59 & 61, 71& 73, 101&103, 107 &109, 137 &139, 149 &151, 179 &181, 191&193, 197 &
TI-83 Programs For Testing Primes and Finding Prime Factors
The TI-83 program PCHECK determines whether a natural number is a prime number. The program PFACT gives the prime factors of a natural number.
PCHECK PFACT : Prompt N : ClrList L 1 : iPart(√ (N)) → M : Prompt N : For(I,2,M) : iPart(√ (N)) → M : If iPart(N/I) = N/I : 2 → I : Then : 1 → J : Goto B : While I ≤ M : End : While N/I ≤ iPart(N/I) : End : I → L 1 (J) : Disp "PRIME" : J+1 → J : Goto C : N/I → N : Lbl B : End : Disp "NOT PRIME" : I+1 → I : Disp I : End : Disp "IS A DIVISOR" : If N > 1 : Lbl C : Then : N → L 1 (J) : End : Disp "STORED IN L 1 " : Disp L 1
Enter these programs into your calculator. The program PCHECK will prompt you for a number. After you input the value it will tell you whether the number
is prime or not. If it is not prime, it will give you a prime divisor of the number. The program PFACT will prompt you for a number, and return a list of prime factors of the number. This list is stored in L 1 for later reference. The screen shot below shows the result of running the PFACT program with an input of
prgmPFACT N=? STORED IN L { 3 3 3 79 } Done
The output tells us that the prime factorization of 2133 is given by (^) ( ) ( ) 3 3 79.
Remark 2.12: PFACT can take considerable time to run if N is large. For example, PFACT takes nearly 30 seconds to give the prime factorization of N = 2303028. Also, it will be necessary to view the list L 1 (using the scrolling arrows) to view the prime factors since the list exceeds the width of the calculator screen.
