# The Fundamental Theorem of Arithmetic

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2 Let us say that any integer n 2 that has the unique factorization property above is a ufp integer. 4 Theorem 1 (The Fundamental Theorem of Arithmetic, or FTA. Every integer n 2 is a ufp integer. 2.2 A proof of the FTC. As indicated above, I will give a proof by contradiction. In order to mae the argument as clear as possible, I will separate out some pieces as claims; some small pieces will be left as exercises. Suppose, in order to get a contradiction, that Theorem 1 were 5 false. Then there would exist at least one integer n 2 that does not have the unique factorization property. Now, if you test all of the integers in order first 2, then 3, then 5, so on to see whether or not each one is a ufp integer, you would in this case eventually arrive at the first non-ufp integer n 0 ; since n 0 would be the first non-ufp integer you had found, it would also clearly be the smallest non-ufp integer. Furthermore, since n 0 would be a non-ufp integer, it would possess two nonidentical prime factorizations n 0 = p 1 p i p r n 0 = q 1 q j q s, where p 1 p i p r q 1 q j q s. The first step in the proof is to establish that the two factorizations in (1 would not only be nonidendical but in fact would have to be completely different, with no factors in common whatsoever: Claim 1.1 In (1, for every 1 i r 1 j s, p i q j ; that is, no p would be equal to any q. Proof (Claim 1.1. Observe first, that by (1, p 1 p r = q 1 q s. (2 Now if, say, p i0 = q j0, we could cancel it from both sides of (2 to get the equality (1 n 1 = p 1 p r p i0 = q 1 q s q j0. (3 Now, as is easy to show, 2 n 1 < n 0 (see Exercise 4. Since 2 n 1, (3 would give two different prime factorizations of n 1. On the other h, all integers in the interval [ 2, n 0 would be ufp integers, so that n 1 would not have two different prime factorizatioms. Since we have arrived at a contradiction, we can conclude that none of the p s could equal any of the q s. (Claim 1.1 We need one particular consequence of Claim 1.1, namely that p 1 q 1 could not be the same prime; 6 without loss of generality, let us say that that p 1 > q 1. Since p 1 > q 1, we can use the two factorizations in (1 to define the positive integer 7 x := ( p 1 p 2 p r ( q1 p 2 p r = (p1 q 1 p 2 p r (4 to ascertain some properties of x. The first property concerns how big or small x could be. 4 Of course, as the FTP says, all integers n 2 are ufp integers, but we cannot assume this fact before we prove it,, until we prove it, this special name will be useful. 5 I will eep the verbs contrary-to-fact to remind you that none of this is actually the case. Proofs by contradiction are not in general written this fastidiously! 6 This is the only consequence that it is safe to use at this point: we do not necessarily have more than one p-prime that is, we cannot rule out the possibility that r = 1 in (2. 7 This definition is the important idea at the center of this proof! 2

3 Claim 1.2 The integer x would have to satisfy the inequalities 2 x < n 0. (5 Proof (Claim 1.2. It follows easily from (4 that x would have to be less than n 0 (see Exercise (1. To prove the assertion that x 2, I will consider two cases. Case 1. If p 1 = 3 in (1, so that q 1 = 2 (why??, then in the factorization n 0 = p 1 p i p r, (6 the index r would need to be at least 2 (see Exercise 2; therefore, starting from Equation (4, we can calculate: x = (p 1 q 1 p 2 p r = (3 2p 2 p r = p 2 p r (because p 2 is really there 2. Case 2. If, on the other h, p 1 3 in (1, then necessarily (p 1 q 1 2 (see Exercise 3, so that x = (p 1 q 1 p 2 p r (2p 2 p r (even if (p 2, p 3,..., p r are not there 2. (Claim 1.2 Claim 1.3 The integer x would have to be a multiple of q 1 ; that is, Proof (Claim 1.3. From (4, we have that x would be the difference q 1 x. (7 x := (p 1 p 2 p r (q 1 p 2 p r. (8 } {{ } } {{ } (α (β Now, the integer (β above is already expressed as a multiple of q 1, while the integer (α by (1 would be n 0, which also would have factorization q 1 q s. Thus, (8 expresses x as the difference of two multiples of q 1 ; this guarantees that x would itself be a multiple of q 1. (Claim 1.3 I am now in a position to assemble the pieces finish the proof. Claim 1.3 says that q 1 x, so that x = q 1 l (9 for some integer l, (9 would be the beginning of a prime factorization of x. On the other h, the right side of (4 would also be the beginning of a prime factorization of x, since 2 x < n 0 (this is Claim 1.2, x would be a ufp integer so that these two factorizations would have to match. In particular: since q 1 : would appear in the prime factorization that started with (9, q 1 would appear in the prime factorization of (p 1 q 1 p 2 p r. (10 Furthermore, since q 1 p i for 2 i r, q 1 would have to show up in the prime factorization of the (p 1 q 1 part of (10. In other words, (p 1 q 1 = q 1 (other primes, 3

4 or, more simply, Finally, adding q 1 to both sides of (11 gives the equation (p 1 q 1 = q 1 (. (11 p 1 = q 1 ( + 1, which would display the prime number p 1 as a product of two integers each of which was 2, contradicting the assumption that p 1 is prime. Exercise 1 Use (4 to show that x would have to be strictly less than n 0. Exercise 2 In (6: show that if p 1 = 3, then r could not equal 1. Exercise 3 Let p 1 q 1 be primes, with p 1 > q 1 p 1 3. Show that (p 1 q 1 2. Exercise 4 Let n 1 be the integer defined in Equation (3. [a]: Show that n 1 < n 0. [b]: Show that 2 n 1. 3 Some Applications. Theorem 1 is called The Fundamental Theorem of Arithmetic because so many of the properties of N Z depend upon it. In this hout, I will discuss a few of the most accessible examples. 3.1 Prime factorizations written using prime powers. Many (most? of the applications of the FTA are most conveniently stated by expressing prime factorizations with multiple copies of the same prime gathered together. For example, instead of writing the prime factorization of 504 as 504 = , one generally writes 504 = or 504 = (12 For some purposes, the form on the left side of (12 is more convenient, while for other applications, 8 the form on the right side of (12 is essential. When two or more integers are being considered at one time, in order to facilitate comparisons among them, it is customary to write each prime factorization so as to have it include every prime that appears in any of the factorizations. For example, if the numbers being considered are , one would generally write 8 Including the famous so-called Gödel numbering. 189 = = , or 189 = =

5 3.2 Divisibility; the gcd; the lcm. The two most basic consequences of the FTA are contained in Exercise 5. Exercise 5 Let n = p a1 1 pa2 2 par r let = p b1 1 pb2 2 pbr r be integers 9 2 with the prime factorizations shown. [a]: Show that [b]: Show that n = p a1+b1 1 p a2+b2 2 p ar+br r. n a 1 b 1 a 2 b 2. a r b r Two further consequences of the FTA concern least common multiples greatest common divisors: Definition 1 Let n be positive integers. [a]: The greatest common divisor of n, denoted gcd(n,, is the largest integer that divides both n. [b]: If gcd(n, = 1, n are said to be relatively prime. [c]: The least common multiple of n, denoted lcm(n,, is the smallest integer that is a multiple of both n. Obviously, for any integer n 1, gcd(1, n = 1 lcm(1, n = n. If n 2 2, Theorem 2 shows how to use the prime factorizations of n 10 to find gcd(n, lcm(n,. In order to state Theorem 2, I find the following notation to be very helpful: for integers 11 a b, let max(a, b be the larger of the two integers let min(a, b be the smaller of the two integers. 12 Theorem 2 Let n = p a1 1 pa2 2 par r Then Proof of Equation (13. Let us put let = p b1 1 pb2 2 pbr r be integers 2 with the prime factorizations shown. lcm(n, = p max(a1,b1 1 p max(a2,b2 2 p max(ar,br r (13 gcd(n, = p min(a1,b1 1 p min(a2,b2 2 p min(ar,br r. (14 M := p max(a1,b1 1 p max(a2,b2 2 p max(ar,br r. To show that M = lcm(n,, I must show that M satisfies the conditions of Definition 1[c]. Step I: Show that n M M. Since a i max(a i, b i for all 1 i r, we can apply Exercise 5[b] to conclude that n M. Similarly, since b i max(a i, b i for all 1 r, we can apply Exercise 5[b] to conclude that M. 9 if one of the numbers equals 1 (say n = 1, the exercise still maes sense is still true. You just let a 1 = = a r = 0 in that case. 10 if the prime factorizations are available. For large integers, prime factorizations are very hard to find, but there is an efficient way to find gcd(n, directly, bypassing prime factorization. 11 They need not be integers; this notation maes sense for all real numbers. However, in this hout, I will be using it only for integers. 12 For the case a = b, obviously, max(a, a = min(a, a = a. 5

6 Step II: Show that M is the smallest among all the common multiples of n. I will actually prove a much stronger statement: I will show that every common multiple of n is a multiple of M. The argument runs as follows. Let C be any common multiple of n. Then, by Exercise 5[b], the prime factorization of C can be written C = p c1 1 pc2 2 pcr r q d1 1 qds s, where the q d1 1 qds s is there to represent whatever primes there are that divide C but do not divide a or b. Now, again by Exercise 5[b], n C = ai c i (1 i r, C = bi c i (1 i r. But then (for each 1 i r, since both a i c i a i c i, obviously max(a i, b i c i. With a final application of Exercise 5[b], we can then conclude that M C. (Equation (13 The proof of Equation (14 is left as Exercise 6. (Theorem 2 Exercise 6 Prove Equation (14. Exercise 7 Use Theorem 2 to prove, for any positive integers n, that 3.3 Recovering bypassed results. gcd(, n lcm(, n = n. As I mentioned in class: the usual way to prove the FTA is to use some preliminary results, which will have been established beforeh. This hout, though, begins by establishing the FTA. We therefore need to derive the results that we have bypassed they are important in their own right. Exercise 8 ass you to derive two of these. Exercise 8 Use the unique factorization property to prove the following important facts about N. [a]: Let n be relatively prime. Show, for any integer l, that if n l then n l. [b]: Let p be prime, let n be integers. Show that if p n, then either p n or p (or both. 3.4 Square roots of positive integers. In class ( in the homewor, we showed that 2 6 are not rational, but we did not answer the full question, Exactly which positive integers have rational square roots? With the FTA, we can now state prove the complete answer this question. I will brea off as a lemma the part of the proof that employs the FTA. 2. Lemma 3 Let n be positive integers. If is a non-whole-number fraction, then so is Proof: Since is a non-whole-number fraction, n is not a multiple of. Therefore, if their prime factorizations are given by n = p a1 1 pa2 2 par r = p b1 1 pb2 2 pbr r, (15 6

7 2: then necessarily b i > a i for at least one 1 i r. Let us suppose that b i0 > a i0. Consider 2 = n2 2 (Why?? = p2a1 1 p 2a2 2 p 2ai 0 i 0 p 2ar r p 2b1 1 p2b2 2 p 2bi 0 i 0 p 2br r 2 furthemore, since b i0 > a i0, obviously 2b i0 > 2a i0. Thus, n 2 is not a multiple of 2, so is not a whole number. Exercise 9 In the proof of Lemma 3: by passing to (15, I was tacitly assuming that n 2 2, this need not be the case. Plug this logical gap: consider the cases n = 1 = 1. Exercise 10 Prove that Lemma 3 remains true even if we drop the assumption that n are positive. Hint: Don t REPROVE Lemma 3; USE it! Theorem 4 A positive integer n has a rational square root if only if n is the square of an integer. Proof ( =: This is completely obvious: if n = 2 for some integer, then obviously n =, a rational number. Proof (= : I will prove the contrapositive statement: If n is not the square of an integer, then n is not rational. Proof of contrapositive: For any positive integer n, there are three logical possibilities for n: (a: n is an integer; (b: n is a non-whole-number fraction; or (c: n is an irrational number. Now, the we are given that n is not the square of an integer, so possibility (a is ruled out; Lemma 3 rules out possibility (b. Thus, possibility (c is the correct one: n is an irrational number. Exercise 11 In the proof of Theorem 4: how exactly does Lemma 3 rule out possibility (b? ; 7

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