University of Miami MTH 210 230 solution manual 5

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Chapter 5 Solutions An Introduction to Mathematical Thinking: Algebra and Number Systems William J. Gilbert and Scott A. Vanstone, Prentice Hall, 2005 Solutions prepared by William J. Gilbert and Alej... andro Morales Exercise 5-1: Verify that the relation ∼ on Z×(Z− {0}), defined by (a, b) ∼ (c, d) if and only if ad = bc, is indeed an equivalence relation. Solution: The relation ∼ on Z × (Z − {0}) is an equivalence relation if it is reflexive, symmetric and transitive. So for (a, b), (c, d) and (e, f) ∈ Z × (Z − {0}) we have (i). ab = ab so (a, b) ∼ (a, b). Therefore ∼ is reflexive. (ii). If (a, b) ∼ (c, d) then ad = bc this is equivalent to cb = da then (c, d) ∼ (a, b). Therefore ∼ is symmetric. (iii) If (a, b) ∼ (c, d) and (c, d) ∼ (e, f) then ad = bc and cf = de. Because b ∈ (Z−{0}), we can divide by b in the first equation to get c = ad b . Substituting for c in the second equation gives ad b f = de. Multiplying both sides by b and dividing by d ∈ (Z − {0}), gives af = be. Hence, (a, b) ∼ (e, f), therefore ∼ is transitive. Then ∼ is indeed an equivalence relation. Exercise 5-2: Verify that addition, subtraction, multiplication and division are all well defined in Q. For example, for addition, it has to be verified that if (a1, b1) ∼ (a2, b2) and (c1, d1) ∼ (c2, d2) then (a1d1 + b1c1, b1d1) ∼ (a2d2 + b2c2, b2d2). Solution: If (a1, b1), (a2, b2), (c1, d1), and (c2, d2) ∈ Z×(Z−{0}) are such that (a1, b1) ∼ (a2, b2) and (c1, d1) ∼ (c2, d2) then a1b2 = a2b1 and c1d2 = c2d1. For addition a1 b1 + c1 d1 = a1d1 + c1b1 b1d1 a2 b2 + c2 d2 = a2d2 + c2b2 b2d2 . Because b1, b2, d1 and d2 ∈ (Z − {0}) then b1d1 and b2d2 ∈ (Z − {0}), and (a1d1 + c1b1)b2d2 = a1b2d1d2 + c1d2b1b2 = a2b1d1d2 + c1d2b1b2 = (a2d2 + c2b2)b1d1. 5.1So (a1d1 + c1b1, b1d1) ∼ (a2d2 + c2b2, b2d2). Hence addition is well defined. For subtraction the same argument works, that is (a1d1 − c1b1)b2d2 = a1b2d1d2 − c1d2b1b2 = a2b1d1d2 − c1d2b1b2 = (a2d2 − c2b2)b1d1. So (a1d1 − c1b1, b1d1) ∼ (a2d2 − c2b2, b2d2). Hence subtraction is well defined. For multiplication a1 b1 · c1 d1 = a1c1 b1d1 a2 b2 · c2 d2 = a2c2 b2d2 . Because b1, b2, d1 and d2 ∈ (Z − {0}) then b1d1 and b2d2 ∈ (Z − {0}), and a1c1b2d2 = a1b2c1d2 = a2b1c2d1 = a2c2b1d1. So (a1c1, b1d1) ∼ (a2c2, b2d2). Hence multiplication is well defined. For division, we assume that dc11 and dc22 6= 0 so c1 and c2 ∈ (Z − {0}), a1 b1 ÷ c1 d1 = a1 · d1 b1 · c1 a2 b2 ÷ c2 d2 = a2d2 b2c2 . Because b1, b2, c1 and c2 ∈ (Z − {0}) then a1c1 and a2c2 ∈ (Z − {0}), and a1d1b2c2 = a1b2d1c2 = a2b1d2c1 = a2d2b1c1. So (a1d1, b1c1) ∼ (a2d2, b2c2). Hence division is well defined. Exercise 5-3: Prove that √3 is irrational. Solution: Suppose that √3 is rational, so there is a rational number x such that x2 = 3. Clearly x 6= 0, so by Proposition 5.11., write x in its lowest terms as a/b where gcd(a, b) = 1. Now (ab )2 = 3 so that a2 = 3b2. Therefore 3|a2 and, since 3 is prime, 3|a. Put a = 3c, so that 9c2 = 3b2 and 3c2 = b2. It now follows that 3|b and so 3 is a common divisor of a and b. This contradicts the hypothesis on a and b being coprime, so it follows that there is no rational number x such that x2 = 3. Exercise 5-4: Prove that √3 4 is irrational. Solution: Suppose that √3 4 is rational. Then √3 4 = a/b where gcd(a, b) = 1. Therefore ab
3 = 4 so a3 = 4b3. Since 4|a3, and 2 is a prime factor of 4, 2|a. Therefore 5.2a = 2c, where c ∈ Z, so (2c)3 = a3 = 4b3, i.e. 2c3 = b3. But this implies that 2|b and thus that 2| gcd(a, b). This contradicts our assumption that gcd(a, b) = 1. Therefore our original assumption, that √3 4 was rational, is false, and √3 4 is irrational. Exercise 5-5: Prove that √6 is irrational. Solution: Suppose that √6 is rational, so there is a rational number r such that r2 = 6. Clearly r 6= 0, so by Proposition 5.11., write r in its lowest terms as a/b where gcd(a, b) = 1. Now (ab )2 = 6 so that a2 = 6b2. Because 3 is a factor of 6 then 3|a2 and, since 3 is prime, 3|a. Put a = 3c, so that 9c2 = 6b2 and 3c2 = 2b2. Then 3|2b2 but because 2 and 3 are coprime then 3|b2. It now follows that 3|b and so 3 is a common divisor of a and b. This contradicts the hypothesis on a and b being coprime, so there is no rational number r such that r2 = 6. Exercise 5-6: Is √2 + √3 rational or irrational? Give reasons. Solution: Suppose √2 + √3 is rational. Then √2 + √3 = m/n (m, n ∈ Z). Square: 2 + √6 + 3 = m2 n2 . Thus √6 = 1/2 m2/n2 − 5
is rational. But √6 is irrational (proof similar to Exercise 5-3). Hence √2 + √3 must be irrational. Exercise 5-7: If a is rational and b is irrational, prove that a + b is irrational. Solution: If there is a rational number r such that r = a+b then because a is rational, so is −a. It follows that r + (−a) = a + b + (−a) = b is also rational as being the sum of two rational numbers. This contradicts the fact that b is irrational, so there is no rational number r such that r = a + b. Exercise 5-8: If a is rational and b is irrational, prove that ab is irrational, except for one case. What is the exceptional case? [Show More]

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