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Lemma 2.8. Let p and q1 , q2 , ..., qn all be primes and let k be a natural number such that pk = q1 q2 · · · qn . Then p = qi for some i. Proof: For the sake of contradiction, assume p 6= qi for any i. Since each qi is prime, p does not divide any of the qi . So p also does not divide q1 · · · qn . Since p does divide pk, it must be the case that pk 6= q1 · · · qn , which is a contradiction. Therefore p must be one of the qi ’s. Exercise 2.10. Express 12! as a product of primes. 12! = 12 ∗ 11 ∗ 10 ∗ 9 ∗ 8 ∗ 7 ∗ 6 ∗ 5 ∗ 4 ∗ 3 ∗ 2 ∗ 1 = (22 ∗ 3) ∗ (11) ∗ (2 ∗ 5) ∗ (32 ) ∗ (23 ) ∗ (7) ∗ (2 ∗ 3) ∗ (5) ∗ (22 ) ∗ (3) ∗ (2) ∗ (1) = 210 ∗ 35 ∗ 52 ∗ 7 ∗ 11 Thm. 2.13. If a and b are natural numbers and a2 |b2 , then a|b. Proof: Express a and b in terms of their prime factorizations: tm a = pr11 · · · prnn and b = q1t1 · · · qm 2t1 2 2rn 2tm 1 Then a2 = p2r 1 · · · pn and b = q1 · · · qm . Since a2 |b2 , all factors of a2 appear as factors of b2 . That is, for each 1 ≤ i ≤ n there exists 1 ≤ j ≤ m such that pi = qj and 2ri ≤ 2tj . Dividing by 2 implies that ri ≤ tj . So each factor of a is also a factor of b, i.e. a|b. Exercise 2.14: gcd(314 ∗722 ∗115 ∗173 , 52 ∗114 ∗138 ∗17) = 30 ∗50 ∗70 ∗114 ∗130 ∗171 = 114 ∗17 Question 2.17: In my opinion, this method of finding the gcd of 2 numbers is more efficient than the Euclidean Algorithm. If we know the prime 1 factorization of the 2 numbers, then there is no question that this method is faster. If we don’t know the prime factorizations, finding them will take approximately the same amount of time as applying the Euclidean algorithm. So, depending on the situation, this method is either much faster or roughly equal to the Euclidean Algorithm. 2