Math 261 Spring 2014 Final Exam May 5, 2014

Transcription

1 Math 261 Spring 2014 Final Exam May 5, Give a statement or the definition for ONE of the following in each category. Circle the letter next to the one you want graded. For an extra good final impression, you can give the definitions/statements for non-circled answers too! (Watch your time though.) (A) The Well-Ordering Axiom. The Well-Ordering Axiom says that every non-empty set of natural numbers has a least element. (B) The Division Algorithm The Division Algorithm says for natural numbers m and n there exist unique q and r such that m = qn + r and 0 r < n. State either one below (or both, for added impression). Circle the letter of the one you want graded. (A) The definition of a prime number. A prime number is a natural number p that cannot be written as p = a b for values a and b that are less than p. (B) The definition of a composite number. A composite number is a natural number n that can be written as n = a b for values a and b that are less than n. State either one below (or both, for added impression). Circle the letter of the one you want graded. (A) The FTOA (Existence) Include all hypotheses and conclusions. The FTOA (Existence) states that every natural number greater than 1 is either prime or can be written as a product of positive powers of primes. In symbols, n > 1 has a prime factorization where p i is prime and r i > 0 for all i. n = p r1 1 prj j, (B) The FTOA (Uniqueness) Include all hypotheses and conclusions. The FTOA (Uniqueness) states that the prime powers in the prime factorization for a natural number are unique and the primes used in the prime factorization are unique up to the order they are written.

2 State either one below (or both, for added impression). Circle the letter of the one you want graded. (A) The definition of a Complete Residue System Modulo n. A Complete Residue System Modulo n is a set of a values {a 1, a 2,..., a k } such that any integer x is congruent modulo n to one of the values in the set. In symbols, given an integer x we must have x a i (mod n) for some a i in our set. (B) The definition of the Canonical Complete Residue System Modulo n. The Canonical Complete Residue System Modulo n is the set {0, 1, 2,..., n 1}. State either one below (or both, for added impression). Circle the letter of the one you want graded. (A) The definition of ord n (a). For natural numbers a and n with (a, n) = 1, ord n (a) is the smallest natural number k such that a k 1(mod n). (B) The definition of the Euler-φ function φ(n). The Euler-φ function φ(n) counts the number of values less than or equal to n > 0 that are relatively prime to n. State either one below (or both, for added impression). Circle the letter of the one you want graded. (A) State Fermat s Little Theorem (any version). Include all hypotheses and conclusions. Fermat s Little Theorem (Version I) says for p a prime and (a, p) = 1, we have a p 1 1(mod p). Equivalently, Fermat s Little Theorem (Version II) says for p a prime and any integer a, we have a p a(mod p). (B) State Euler s Theorem. Include all hypotheses and conclusions. Euler s Theorem says for n > 0 and (a, n) = 1, we have a φ(n) 1(mod n).

3 2. Make a calculation (with some supporting justification) for ONE of the following in each category. Circle the letter next to the one you want graded. For an extra good final impression, you can make the other calculations too! (Watch your time though.) (A) Find x and y to write 96x + 21y = (96, 21). Begin by using the Euclidean Algorithm to find (96, 21): 96 = = = = So (96, 21) = 3. Then using the 3rd equation we know (96, 21) = 3 = Rewriting the 2nd equation gives 9 = and this combines with the previous equation to give (96, 21) = 12 1 ( ) = Finally, rewriting the 1st equation gives 12 = Combining this with the previous equation (96, 21) = = 2 ( ) 1 21 = So x = 2 and y = 9. (B) Find the number of zeroes at the end of the normal (base ten) representation of 24!. By the FTOA, every factor in 24! can be written as a product of primes. In particular, we are interested in the number of times the primes 2 and 5 appear since a zero at the end of a number corresponds to a factor of 10 = 2 5. Note that every other (even) factor in 24! contributes at least one power of 2, with 4 = 2 2, 8 = 2 3, and 12 = 2 2 3, 16 = 2 4, 20 = and 24 = contributing more. So there are = 2 20 powers of 2 in the prime factorization of 24!. Only every fifth factor in 24! contributes a power of 5, and these come from 5, 10 = 5 2, 15 = 5 3, and 20 = 5 4. So there are = 5 4 powers of 5 in the prime factorization of 24!. Each pair of 2 and 5 (multiplied together) contributes one more zero at the end of 15!. Since there are only four such pairs (because there are only four factors of 5), there must be FOUR zeroes at the end of 24!.

4 Question 2 continued... Calculate either one below (or both, for added impression). Circle the letter of the one you want graded. (A) Find the CCRS value corresponding to 2 67 (mod 31). By Fermat s Little Theorem (Version I), since (2, 31) = 1 we have Since 67 = we must have (mod 31) = = (2 30 ) (mod 31). It is easy to calculate 2 7 = 128. The correct CRS value a is such that 128 a is divisible by 31. Note that 4 31 = 124. So = 124 = 4 31 is divisible by 31. Therefore (mod 31). We could also perform this calculation by noting that ord 31 (2) = 5 since 2 5 1(mod 31). (B) Find any CRS value that is the inverse of 5(mod 57). First note that 57 = 3 19 is a composite number. Since (5, 57) = 1, Euler s Theorem says 5 φ(57) 1(mod 57). From our work in class, φ(57) = φ(3) φ(19) (since 3 and 19 are relatively prime) and it follows that φ(57) = 2 18 = 36. Then (mod 57). Then b = 5 35 is a CRS value that is the inverse since = (mod 57).

5 3. Let a, b, and n be integers with n > 0. The equation ax b(mod n) has a solution if and only if (a, n) b. You may cite only Chapter 1 results without proof. Proof: We will break this proof into two parts corresponding to the if and only if. Suppose ax b(mod n) has a solution. This means there exists a value for x such that n (ax b). By definition of divides, there is an integer z such that nz = ax b. Then ax nz = b. Taking y = z gives integers x and y with ax + ny = b. Now by Chapter 1 results, this equation only has solutions if (a, n) b. Now suppose (a, n) b with b = (a, n) k for some integer k. By Chapter 1 results, aˆx + nŷ = (a, n) for some integers ˆx and ŷ. Then aˆxk + nŷk = (a, n) k = b. Now nŷk = aˆxk b or n (aˆxk b). Then by definition of congruence aˆxk b(mod n) gives a solution to ax b(mod n) where x = ˆxk.

6 4. Prove ONE of the following statements (or both, for added impression). Clearly indicate which proof you are writing. If you write both, put one on the reverse side: (A) Show that there are no two natural numbers a and b such that a2 = 3. (This is a necessary step b2 to prove 3 is irrational, but you don t have to prove this latter statement) Proof: Assume for a contradiction that a2 b 2 = 3. This implies that a2 = 3b 2. By FTOA, both a and b have prime factorizations, say a = p r1 1 prj j by algebra a 2 and b 2 have prime factorizations a 2 = p 2r1 1 p 2rj j and b 2 = q 2s1 1 q 2r k k. and b = q s1 1 qr k k. Then Since each exponent of the primes in the factorization of a 2 is multiplied by 2, every power must be even. Similarly, each exponent of the primes in the factorization of b 2 is multiplied by 2, every power must be even. However, a 2 = 3b 2 implies that the power of 3 in the prime factorization of a 2 must be odd, since 3b 2 gives an odd (even plus one) number of occurrences of 3. This is contradicts that FTOA says prime factorizations are unique, but earlier we saw all powers in the prime factorization of a 2 were even. Therefore, we cannot have two natural numbers a and b such that a2 b 2 = 3. (B) Assume a2 = 3 for some natural numbers a and b. Then consider the set b2 S = {k 3 k and k 3 are natural numbers}. What can you say about this set? Try to use this set to prove 3 is irrational. Proof: Assume a2 = 3 for some natural numbers a and b and consider the set S. b2 Since a2 b 2 = 3 we have a b = 3 and a = b 3. Since a is a natural number, b 3 must be a natural number. This implies that S is a non-empty set of natural numbers. By WOA, S must have a least element, call it l. We can show 3 is irrational by demonstrating our assumptions lead to a contradiction. Consider our least element l. Since l S, there must be some natural number k with l = k 3. It follows that l 3 = k 3 3 = 3k is also a natural number. Now check this out: l 3 l = l( 3 1) = 3k l must also be a natural number. Also ( 3 1) > ( 1 1) = 0 and 1 = ( 4 1) > ( 3 1). But then l( 3 1) is a natural number less than l. Multiplying l( 3 1) by 3 gives 3l 3l which, by supposition, is the natural number 3l 3k. These imply that l( 3 1) is in S, but this contradicts our choice of l as the least element. Therefore, we cannot have natural numbers a and b with a2 b 2 = 3. 6

7 5. Prove ONE of the following statements (or both, for added impression). Clearly indicate which proof you are writing. If you write both, put one on the reverse side: (A) If p and q are distinct primes, then p q 1 + q p 1 = 1(mod p q). Proof: By Fermat s Little Theorem (Version I) p q 1 1(mod q) and q p 1 1(mod p). Note since p 0(mod p) and q 0(mod q) we also have p q 1 0 q 1 (mod p) 0(mod p) and q p 1 0 p 1 (mod q) 0(mod q). Putting together the pieces modulo p and q now gives: p q 1 + q p (mod p) 1(mod p) p q 1 + q p (mod q) 1(mod q) Now by Theorem 4.21 above, we have p q 1 + q p 1 1(mod p q). (B) If p is prime and a is an integer with 1 < a < p then p (a p + (p 1)!a). Proof: We must show that p (a p + (p 1)!a) or that a p + (p 1)!a = k p for some k Z. By Fermat s Little theorem, we know that a p a (mod p). By Wilson s Theorem, (p 1)! 1 (mod p), and it follows (using a Ch. 1 result) that a(p 1)! a (mod p). Then a p + (p 1)!a a + (p 1)!a (mod p) By FLT and Chapter 1 results a + ( a) (mod p) From what follows after Wilson s Theorem 0 (mod p) But this says that a p + (p 1)!a 0 = k p for some k Z. So p (a p + (p 1)!a).

8 6. Show n 5 n is divisible by 5 for every natural number n by writing a proof by induction. If you cannot prove it using induction, prove it another way (for reduced credit). Proof: Begin by considering the base case when n = 1. Since = 0 = 0 5, we see that 5 (1 2 1). Assume, as our induction hypothesis that the result holds up to n. That is, for every value i less than or equal to n we have 5 (i 5 i). In particular 5 (n 5 n) and n 5 n = 5j for some integer j. We now need to consider n + 1. Note that (n + 1) 5 (n + 1) can be written as n 5 + 5n n n 2 + 5n + 1 (n + 1) = n 5 n + 5(n 4 + 2n 3 + 2n 2 + n). By our induction hypothesis, 5 (n 5 n) and n 5 n = 5j. Then n + 1) 5 (n + 1) = 5j + 5(n 4 + 2n 3 + 2n 2 + n) = 5(j + n 4 + 2n 3 + 2n 2 + n). This shows that 5 ((n + 1) 5 (n + 1)) and the desired conclusion follows from proof by induction. Alternate, Reduced-Credit Proof: We know that p = 5 is a prime. Then by Fermat s Little Theorem (Version II) a 5 a(mod 5) for all natural numbers a. By definition of congruence, 5 (a 5 a) for such values of a. If instead we use Fermat s Little Theorem (Version I) then for (a, 5) = 1 we have a 4 1(mod 5). Multiplying this equation by a now gives a 5 a(mod 5) and the result follows as above. For a not relatively prime to 5, we must have 5 a since 5 is a prime. Then a 0(mod 5) and a (mod 5). By Chapter 1 results, a 5 a 0 0(mod 5) 0(mod 5). Then by definition of congruence, this says that 5 (a 5 a 0) or 5 (a 5 a).