Prime Numbers
A Prime number is a natural number with exactly two distinct natural number divisors: 1 and itself. Otherwise a number is called a composite number. Therefore, 1 is not a prime, since it only has one divisor, namely 1. A number \(n > 1\) is prime if it cannot be written as a product of two factors \(a\) and \(b\), both of which are greater than 1: n = ab.
• The first twenty-six prime numbers are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101
• Note: only positive numbers can be primes.
• There are infinitely many prime numbers.
• The only even prime number is 2, since any larger even number is divisible by 2. Also 2 is the smallest prime.
• All prime numbers except 2 and 5 end in 1, 3, 7 or 9, since numbers ending in 0, 2, 4, 6 or 8 are multiples of 2 and numbers ending in 0 or 5 are multiples of 5. Similarly, all prime numbers above 3 are of the form \(6n-1\) or \(6n+1\), because all other numbers are divisible by 2 or 3.
• Any nonzero natural number \(n\) can be factored into primes, written as a product of primes or powers of primes. Moreover, this factorization is unique except for a possible reordering of the factors.
• Prime factorization: every positive integer greater than 1 can be written as a product of one or more prime integers in a way which is unique. For instance integer \(n\) with three unique prime factors \(a\), \(b\), and \(c\) can be expressed as \(n=a^p*b^q*c^r\), where \(p\), \(q\), and \(r\) are powers of \(a\), \(b\), and \(c\), respectively and are \(\geq1\).
Example: \(4200=2^3*3*5^2*7\).
• Verifying the primality (checking whether the number is a prime) of a given number \(n\) can be done by trial division, that is to say dividing \(n\) by all integer numbers smaller than \(\sqrt{n}\), thereby checking whether \(n\) is a multiple of \(m\leq{\sqrt{n}}\).
Example: Verifying the primality of \(161\): \(\sqrt{161}\) is little less than \(13\), from integers from \(2\) to \(13\), \(161\) is divisible by \(7\), hence \(161\) is not prime.
Note that, it is only necessary to try dividing by prime numbers up to \(\sqrt{n}\), since if n has any divisors at all (besides 1 and n), then it must have a prime divisor.
• If \(n\) is a positive integer greater than 1, then there is always a prime number \(p\) with \(n < p < 2n\).
https://gre.myprepclub.com/forum/gre-qu ... tml#p51913From our quant book also
I think you have misunderstood something................
_________________