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Math 4310 Solutions to homework 1 Due 9/1/16 1. Use the Euclidean algorithm to find the following greatest common divisors. (a) gcd(252, 180) = 36 252 = 180 · 1 + 72 180 = 72 · 2 + 36 72 = 36 · 2 + 0 (b) gcd(513, 187) = 1 513 = 187 · 2 + 139 187 = 139 · 1 + 48 139 = 48 · 2 + 43 48 = 43 · 1 + 5 43 = 5 · 8 + 3 5=3·1+2 3=2·1+1 2=1·2+0 (c) gcd(7684, 4148) = 68 7684 = 4148 · 1 + 3536 4148 = 3536 · 1 + 612 3536 = 612 · 5 + 476 612 = 476 · 1 + 136 476 = 136 · 3 + 68 136 = 68 · 2 + 0 2. Find the multiplicative inverses of the given elements, or explain why there is not one. (a) [38]−1 = [59] in Z83 First we do the Euclidean algorithm: 83 = 38 · 2 + 7 38 = 7 · 5 + 3 7 = 3 · 2 + 1. Substituting backwards, 1 = 7 − 3 · 2 = 7 − (38 − 7 · 5) · 2 = 7 · 11 − 38 · 2 = (83 − 38 · 2) · 11 − 38 · 2 = 83 · 11 − 38 · 24. Hence [38]−1 = [−24] = [59] in Z83 . Math 4310 (Fall 2016) Solution 1 2 (b) [351] has no inverse in Z6669 because [351] · [19] = [0]. (c) [91]−1 = [451] in Z2565 First we do the Euclidean algorithm: 2565 = 91 · 28 + 17 91 = 17 · 5 + 6 17 = 6 · 2 + 5 6 = 5 · 1 + 1. Substituting backwards, 1 = 6 − 5 · 1 = 6 − (17 − 6 · 2) · 1 = 6 · 3 − 17 · 1 = (91 − 17 · 5) · 3 − 17 · 1 = 91 · 3 − 17 · 16 = 91 · 3 − (2565 − 91 · 28) · 16 = 91 · 451 − 2565 · 16. Hence [91]−1 = [451] in Z2565 . 3. Prove the following facts about divisors. In the following, a, b, c ∈ N. (a) If b|a, then b|ac. Since b|a, there is a natural number k such that bk = a. Multiplication is well-defined, so (bk)c = ac. Multiplication is also associative, so b(kc) = ac. Finally, since N is closed under multiplication, kc ∈ N, and by definition, b|ac. (b) If b|a and c|b, then c|a. Since c|b, there is a natural number k1 such that ck1 = b. (1) Since b|a, there is a natural number k2 such that bk2 = a. (2) We can multiply both sides of (1) by k2 to get (ck1 )k2 = bk2 . Multiplication is associative, so c(k1 k2 ) = bk2 . Equality is transitive, so using (2), we have c(k1 k2 ) = a. Finally, N is closed under multiplication, so k1 k2 ∈ N, and by definition c|a. (c) If c|a and c|b, then c|(ma + nb) for any integers m, n. Since c|a and c|b, there are natural numbers k1 and k2 such that ck1 = a, ck2 = b. Multiplication is well-defined and associative, so c(k1 m) = am, c(k2 n) = bn. Multiplication is also commutative, so c(k1 m) = ma, c(k2 n) = nb. Math 4310 (Fall 2016) Solution 1 3 Addition is well-defined, so c(k1 m) + c(k2 n) = ma + nb. Distributivity gives c(k1 m + k2 n) = ma + nb. Finally N is closed under multiplication and addition, so k1 m + k2 n ∈ N, and by definition c|(ma + nb). 4. Fill in the addition and multiplication tables for F5 . + 0 1 2 3 4 · mod 5 0 1 2 3 4 0 0 1 2 3 4 0 0 0 0 0 0 1 1 2 3 4 0 1 0 1 2 3 4 2 2 3 4 0 1 2 0 2 4 1 3 3 3 4 0 1 2 3 0 3 1 4 2 4 4 0 1 2 3 4 0 4 3 2 1 mod 5 (a) How can we conclude from the addition and multiplication tables that the operations are commutative? The tables are symmetric across the top-left-to-bottom-right diagonal, so the operations are commutative. (b) How can we conclude that 0 is an additive identity? The first row of the addition table shows that 0 + a = a for all a, while the first column of the same table shows that a + 0 = a for all a. This suffices to show that 0 is an additive identity. (c) How can we conclude that every non-zero element has a multiplicative inverse? Excluding the 0 row, each row in the multiplication table has a 1, showing that for every a, there is a b such that ab = 1. Excluding the 0 column, each column in the multiplication table has a 1, showing that for every a, there is a b such that ba = 1. Hence every non-zero element has a left and right multiplicative inverse. (d) Working with elements in F5 , find the solution to the linear equation 3x + 2 = 4. Adding 3 to both sides, we get 3x = 2. Multiplying both sides by 2, we get x = 4. 5. Taking for granted that R is a field, we can verify the field axioms for C. Recall that for two complex numbers z = a + bi and w = c + di, with a, b, c, d ∈ R, we have z + w = (a + c) + (b + d)i and z · w = (ac − bd) + (ad + bc)i. (a) Prove that + is associative. (Sorry, it’s messy! You have to do this kind of thing once. Also, associativity of · is worse.) Math 4310 (Fall 2016) Solution 1 4 Our three complex numbers will be zk = ak + bk i for k = 1, 2, 3. Using the definition of addition, (z1 + z2 ) + z3 = ((a1 + b1 i) + (a2 + b2 i)) + (a3 + b3 i) = ((a1 + a2 ) + (b1 + b2 )i) + (a3 + b3 i) = ((a1 + a2 ) + a3 ) + ((b1 + b2 ) + b3 )i Since addition for R is associative, the last line becomes ((a1 + a2 ) + a3 ) + ((b1 + b2 ) + b3 )i = (a1 + (a2 + a3 )) + (b1 + (b2 + b3 ))i. Use the definition of addition again but now in the opposite direction: (z1 + z2 ) + z3 = (a1 + (a2 + a3 )) + (b1 + (b2 + b3 ))i = (a1 + b1 i) + ((a2 + a3 ) + (b2 + b3 )i) = (a1 + b1 i) + ((a2 + b2 i) + (a3 + b3 i)) = z1 + (z2 + z3 ), as desired. (b) Suppose z = a + bi satisfies a 6= 0 and b 6= 0. What is the multiplicative inverse of z? a b Let y = a2 +b 2 − a2 +b2 i. Because a and b are both nonzero, their squares are both positive, so a2 + b2 > 0. Thus y is well-defined. Now we check the products: b a − i zy = (a + bi) · a2 + b2 a2 + b2 a2 ab ba b2 = 2 − i + i − i2 . a + b2 a2 + b2 a2 + b2 a2 + b2 Because i2 = −1 and multiplication is commutative in R, this becomes zy = a2 ab ab b2 a2 b2 − i + i + = + 0 + = 1. a2 + b2 a2 + b2 a2 + b2 a2 + b2 a2 + b2 a2 + b2 This shows that y is a right inverse of z. We check the other: a b − i · (a + bi) yz = a2 + b2 a2 + b2 a2 ab ba b2 = 2 + i − i − i2 . a + b2 a2 + b2 a2 + b2 a2 + b2 Because i2 = −1 and multiplication is commutative in R, this becomes yz = a2 ab ab b2 a2 b2 + i − i + = + 0 + = 1. a2 + b2 a2 + b2 a2 + b2 a2 + b2 a2 + b2 a2 + b2 This shows that y is a left inverse of z. Thus y is the multiplicative inverse of z. Math 4310 (Fall 2016) Solution 1 5 (c) In part (b), if a, b ∈ Q, is z−1 ∈ Q[i]? Because Q is closed under multiplication and addition, if a, b are in Q, then so is a2 +b2 . a −b As argued before, a2 +b2 6= 0, and Q is a field, so (a2 +b2 )−1 ∈ Q. Then a2 +b 2 , a2 +b2 ∈ Q, a −b so z−1 = a2 +b 2 + a2 +b2 i is in Q[i]. Extended Glossary. Please give a definition of a prime number. Then give an example of a prime number, an example of a number that is not a prime number (don’t forget to explain why!), and state and prove a theorem about prime numbers. In studying multiplication in the natural numbers, we quickly run into the idea of expressing numbers as products of smaller numbers. Since 1 is the multiplicative identity, it is most interesting when the components are not 1. For instance 12 = 4 · 3, where both 4 and 3 are smaller than 6 and neither is 1. Eventually we cannot break numbers down further, giving us our “minimal” units of multiplication, which we call prime numbers. Definition 1. A prime number is a natural number greater than 1 with no positive divisors except 1 and itself. Example 2. 3 is a prime number. Any divisors of 3 must be smaller than 3, which means the only possible nontrivial (i.e. neither 1 nor 3) divisor is 2. However 2 does not divide 3, so 3 is prime. Example 3. 12 is not prime because 12 is divisible by 2. A natural question is, how many prime numbers are there? We will show that there are infinitely many of them, but before we can do that, we need the following lemma: Lemma 4. Every natural number greater than 1 has a prime divisor. Proof. Let S be the set of natural numbers greater than 1 which have no prime divisors, and suppose S is nonempty. Then by the Well-Ordering Principle, S has a least element `. Since ` has no prime divisors, in particular ` is not a prime number. We defined S such that ` > 1, so ` has a divisor which is neither 1 nor itself, which we will call d. We have that d divides ` but 1 < d < `, so d must have a prime divisor p, since it is not in S. However if p divides d and d divides `, then p divides `, which contradicts our construction of `. Hence our initial assumption must be incorrect, and S is empty. Therefore every natural number greater than 1 has a prime divisor. Theorem 5. There are infinitely many prime numbers in N. Proof. Suppose there are only finitely many prime numbers in N, and they are a1 , a2 , . . . , an . Let b = a1 a2 · · · an + 1. Since N is closed under addition and multiplication, b is in N. By lemma 4, every natural number greater than 1 has a prime divisor, so b has a prime divisor, which must be ai for some i. Since ai divides a1 a2 · · · an and ai divides a1 a2 · · · an + 1, ai must divide 1. However 1 has no prime divisors, so our initial assumption was wrong, and there are infinitely many primes in N.