Math 126 Number Theory

Prof. D. Joyce, Clark University 3 Feb 2006

Due Monday. From page 32, exercises 5, 7, 12, Theorem. Suppose that a is not a perfect nth power, 13, 15. that is, that there is no b such that bn = a. Then th Due Wednesday. From page 35, exercises 1, 2. the n root of a is not a rational number. Proof: We’ll prove the contrapositive instead which For next time. Read through section 2.4. √ says that if n a is a rational number, then a is a Last meeting. The unique factorization theorem. perfect nth power. √ n Today. Some applications of the unique factoriza- √ Suppose that a is a rational number, that is, tion theorem including irrationality of surds. Some n a = r/s where r and s are integers. We may of these theorems are interesting, like the irrational- assume that r and s are relatively prime since oth- ity of surds, some are clarifying, like the next one, erwise we could divide them both by their greatest n n and some some might be called lemmas since their common divisor (r, s). Then a = r /s , and so purpose won’t be clear until we use them. n n The ﬁrst one just clariﬁes what the prime factor- as = r . izations of powers look like. We’ll show that s = 1. Theorem. The exponents in the prime factorization Suppose that s > 1. Then some prime p divides of an nth power cn of a number c are all divisible by s. Since p|asn, therefore p|rn. Therefore, by the n. Thus, if a prime p divides cn, then its nth power previous theorem, p|r. But p can’t divide both r also divides cn, so p divides c. and s since they’re relatively prime, a contradic- Proof: Let the prime factorization of c be tion. Therefore s = 1. n e1 e2 er Since s = 1, therefore a = r . q.e.d. c = p1 p2 ··· pr . The following theorem isn’t particularly interest- Then a prime factorization of cn is ing, but it will be useful later when we solve Dio- phantine equations. cn = pne1 pne2 ··· pner . 1 2 r Lemma. If an nth power is the product of two rela- The unique factorization theorem says that there tively prime numbers, then each of those number is only one prime factorization, and in this prime are also nth powers. That is, if cn = ab where factorization the exponents are all divisible by n, (a, b) = 1, then there exist numbers d and e such therefore the exponents in the prime factorization that dn = a, en = b, and c = de. n of c are all divisible by n. q.e.d. Proof: Since a and b are relatively prime, the primes The next theorem is interesting.√ It’s a general that appear in their prime factorizations are dis- theorem that shows that surds like 2 are irrational tinct. So if numbers. A surd is the nth root of a number that th e1 e2 er er+1 er+s is not a perfect n power. a = p1 p2 ··· pr and b = pr+1 ··· pr+s

1 are the prime factorizations of a and b where the n pi’s are distinct. Then c = ab has the prime factorization

n e1 e2 er er+1 er+s c = p1 p2 ··· pr pr+1 ··· pr+s .

Now, since cn is a perfect nth power, all the exponents ei in this prime factorization are divisible by n. That means all the exponents in the prime factorizations of a and b are divisible by n. Hence, a and b are nth powers. q.e.d.