THE FUNDAMENTAL THEOREM OF ARITHMETIC ( 1) Every integer N 2 has exactly one prime factorization.

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Gertrude Joseph
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THE FUNDAMENTAL THEOREM OF ARITHMETIC ( 1) PROLOGUE This is an amazing theorem. It makes a not especially apparent statement about an infinite number of numbers, and has many neat applications.

1 THE FUNDAMENTAL THEOREM OF ARITHMETIC ( 1) PROLOGUE This is an amazing theorem. It makes a not especially apparent statement about an infinite number of numbers, and has many neat applications. The deeper you probe mathematics, the more amazing it becomes. At each level there are new, unprecedented connections. Mathematics is not just boringly logical; it is extravagantly wonderful. It seems to go beyond what is required just to keep the universe together. Why is that the case? Christians know why: God made mathematics. It reflects His infinite creative power and His joy in creation. He made many incredible things that have only been just recently discovered, after thousands of years; probably He has made many wonders, like stars, that man will never discover. God does not depend on our praise. But praise is our proper response to creation, as we find in Psalm 8. We study mathematics so that we may know more of God s wonderful and majestic handiwork, so that our soul may sing to Him. Sadly, some have perverted the study of mathematics, as of the rest of God s creation, so that they worship the creation itself or even their own intelligence in discovering it. This response is completely wrong; but it is still easy to fall into. So, let us consciously avoid error and praise God for what He has made! Below follow some preliminary theorems and a definition necessary to the proof. THEOREM 1 Existence of a Prime Factorization for All Real Integers N 2 Every positive integer N 2 has at least one representation as a multiplication of primes ( prime factorization ). If N is a prime, then it is its own prime factorization. If N is not a prime, then (by definition) it can be written as the product of two integers greater than 1. Both of these numbers in turn must be either a prime or the product of two integers greater than 1. This process can be represented by an organizational tree. For example, N Factor 1 Factor 2 (prime) Factor 1,1 (prime) Factor 1,2 (prime) In this case, the prime factorization of N is (Factor 1,1 Factor 1,2 Factor 2 ). The above process may be protracted as long as necessary. Each factor found is either a prime factor of N or may be divided into new factors. Since N is finite, it cannot be the product of an infinite number of integers greater than 1, so this algorithm must stop at some point;

2 THE FUNDAMENTAL THEOREM OF ARITHMETIC ( 2) when it does, we simply read across the bottom of the tree as above to find a prime factorization of N, thus implying that one exists. DEFINITION: A quotient of positive integers is in lowest terms iff no positive integer j exists such that and are both positive integers. THEOREM 2 If the numerator and denominator of a fraction a/b are uniquely factored, and every factor of its numerator or denominator is uniquely factored, then it can be uniquely reduced to lowest terms. Let the set C = {P 1, P 1, P k } be the set of common prime factors of the numerator a and the denominator b. C includes multiplicities; for example, 2 would be included twice for the fraction 12 / 56 = / Let F = the product of all the primes in C. Then, for two integers l and m, a = lf and b = mf. l and m are uniquely factored, by the condition of the theorem. So they must not have any common factors such factors would be common factors of a and b, and are in C. a/b in lowest terms is equivalent to, which cannot be reduced any more. l and m are uniquely determined, since the members of C are uniquely determined by the condition of the theorem. PROVING THE FUNDAMENTAL THEOREM OF ARITHMETIC The Fundamental Theorem of Arithmetic: Let N be an integer greater than 1. Then N has exactly one factorization of primes. By Theorem 1, N has at least one factorization of primes. Our strategy is to show that all prime factorizations of N must be identical, by proving that one factorization (F 1 ) cannot have a prime different than another factorization (F 2 ). S(n) is that any prime factorization of length n equaling N is the only prime factorization of any length of N. We will use mathematical induction, proving S(1) and then S(k+1) given S(1) through S(k). S(1) In this case, F 1 = P = N, where P is a prime number. By definition, the only factorization (prime or otherwise) of P is P, and this holds for N also, since N = P. Proving S(k+1) given S(1) S(k)

3 THE FUNDAMENTAL THEOREM OF ARITHMETIC ( 3) Let F 1 = P 1 P 2 P k+1 = N. Let F 2 = R 1 R 2 R j be some other prime factorization of N, where all of P and R are prime. (We here observe that j k+1, because otherwise we have proven S(j), making F 2 the only prime factorization of N, which is impossible since F 1 is a prime factorization of N.) We know that N = F 1 = F 2. So, and. We now seek to show that the above statement can only be true if each prime factor of F 2 corresponds to an equivalent prime factor of F 1, and vice versa, making them the same prime factorizations of N. If in fact, then, (Equation 1) (We will refer to the expression on the left side of the Equation 1 as Q 1 and the expression on the right as Q 2 for the remainder of this proof; their numerators and denominators will be denoted respectively as Q N1, Q N2, Q D1, and Q D2.) We now conclude the proof by showing that in either of the two possible states of Q 1, reducible and fully reduced, F 1 and F 2 are the same prime factorizations of N. If Q 1 can be reduced, that is, its numerator R k+1 R k+2 R j is evenly divisible by P 1, then its denominator, P 1, cancels with the numerator. In that case Q 1 is an integer, and so also Q 2 ; in this case, F 1 and F 2 are the same factorizations of N, as shown below: Neither the numerator nor the denominator of Q 2 can be refactored by S(k). Thus, by Theorem 2, it is possible to uniquely reduce Q 2 to lowest terms by canceling all common primes of its numerator and denominator. Is Q 2, fully reduced, an integer? If so, then each prime term R 1 R k in its denominator must cancel with a prime term P 2 P k+1 of its numerator. (Note that we can say this with surety because, again, neither the numerator nor the denominator of Q 2 can be refactored.) Because the numerator and denominator of Q 2 are composed of the same number of primes, Q 2, if it is an integer, equals 1/1 = 1. In this case, to preserve the equality, R k+1 must equal P 1 and F 1 and F 2 are the same factorizations of N, since each P i corresponds exactly to an R t, and vice-versa. If Q 1 cannot be reduced: First of all, we show that in this case at least one of the members of F 2 {R 1, R 2, R j } equals P 1. We know that there exists some constant fraction r = t/s such that rq N1 = Q N2 and rq D1 = Q D2.

4 THE FUNDAMENTAL THEOREM OF ARITHMETIC ( 4) Take two equivalent fractions a/b = c/d. Then the ratio r of their numerators is c/a. Since (c/a) / (c/a) = 1 and ar = c, (c/d) = (a/b) = (a/b) (r/r) = c/(br). By the transitive law of equality, c/d = c/(br). Thus, ar = c and br = d for some fraction r. continued on next page Furthermore, r must be an integer, since Q 1 is fully reduced. Take two equivalent fractions a/b = c/d such that a/b is fully reduced and ar = c and br = d. Let r, fully reduced, equal t/s. Then a and b must be evenly divisible by s, since c and d are integers (s cannot cancel with a factor of t because t and s have no common prime factors by the definition of irreducibility). However, since a/b is fully reduced and hence a and b have no common prime factors s must equal 1. Thus, r = t/s = t, an integer. So, we know that and thus Now the product Q D2 = R 1 R 2 R k is uniquely factored, by S(k). Hence all the primes by which it is evenly divisible are in the set {R 1, R 2, R k }. Since Q D1 divides evenly into Q D2, Q D1 = P 1 is a member of the set {R 1, R 2, R k }. So, P 1 is equivalent to some R i (1 i k). For convenience, we let i=1. (F 2 can be commutatively altered without changing any of its factors.) So,. Now, we seek to show using mathematical induction that, for each i from 1 to k+1,. T(n) is that.we have proven T(1). The logic differs somewhat for T(k) and T(k+1), so they are proven subsequent to the following. Proving T(n) for n < k (given P i = R i for 1 i<n): Recall Equation 1:. Also recall that the numerator Q N2 nor the denominator Q D2 of Q 2 can be refactored by S(k). We know that Q N2 may be evenly divided by the product D = R 2 R 3 R n-1. Simplifying:

5 THE FUNDAMENTAL THEOREM OF ARITHMETIC ( 5) Now, multiply both sides of the above equality by : Simplifying (remember that R 1 = P 1, and so the two terms cancel), Now: is Q 1 in lowest terms, or not? If it is not in lowest terms, then the logic applied to Equation 1 still holds for this equation (Q N2 and Q D2 being uniquely factored), the remaining factors of F 1 and F 2 cancel with each other, and F 1 and F 2 are the same prime factorizations of N. If Q 1 is in lowest terms, then (as shown above) P n is a member of the set, let us say R n. Thus. Proving T(k) and T(k+1) (given P i = R i for 1 i<k) Again recall Equation 1: We know that Q N2 may be evenly divided by the product D = R 2 R 3 R k-1. Simplifying: Now, multiply both sides of the above equality by : Simplifying (remember that R 1 = P 1, and so the two terms cancel), Now Q 2 is unquestionably in lowest terms, by S(1). So, for some integer r,. But P k is a prime number, and can have only one prime factor, so r must equal 1. Thus,, and, since r=1,. This proves T(i) for 1 i k+1. Thus F 1 and F 2 are the same prime factorizations of N; this concludes the proof of S(k+1). Q.E.D. Hallelujah!

6 THE FUNDAMENTAL THEOREM OF ARITHMETIC ( 6)

A field F is a set of numbers that includes the two numbers 0 and 1 and satisfies the properties:

Byte multiplication 1 Field arithmetic A field F is a set of numbers that includes the two numbers 0 and 1 and satisfies the properties: F is an abelian group under addition, meaning - F is closed under