1 Mathematics for Liberal Arts c 2008 Robert O. Stanton 1.1 Module 7 We have been reducing fractions to lowest terms since elementary school. We now put this procedure on a sound theoretical footing. 4 12 20 6 2 , , , . All of these reduce to . The 6 18 30 9 3 following de…nition tells what it means for two rational numbers to be equivalent. Consider the rational numbers De…nition. Let a; b; c; d be integers, b 6= 0, d 6= 0. The rational numbers c are equivalent provided that ad = bc. d a and b 12 4 We check on a couple of the examples above: say and :Then 4 18 = 72 and 6 18 20 6 12 6 = 72. For and , we have 20 ( 9) = 180 and 6 30 = 180. We can 30 9 check other pairs as well. The following theorem provides a relationship between equivalent rational numbers. c a Theorem. If a; b; c; d are integers, b 6= 0; d 6= 0, and is equivalent to , then b d there is a rational number r such that ar = c and br = d. Proof. B: there is a rational number r A: a; b; c; d are integers, b 6= 0; d 6= 0, c a such that ar = c and br = d. and is equivalent to b d start with forward. A1 . ad = bc ad A2 . =c b d A3 . a = c. b d A4 . Let r = . Then r is a rational number b and ar = c. d = d Done. A5 . br = b b 1 We have been reducing fractions to lowest terms since elementary school. But what does it mean for a fraction to be in lowest terms? The following de…nition makes this concept precise. De…nition. Let a and b be integers with b 6= 0. Then the rational number in lowest terms provided that b > 0 and gcd (a; b) = 1. a is b We now show that it is possible to put an integer in lowest terms. For simplicity, we restrict consideration to positive integers. a Theorem. If a and b are positive integers, then the rational number is equivb alent to a rational number in lowest terms. Proof. a B: the rational number is equivalent A: a and b are positive integers b A3 . Let d = gcd (a; b). to a rational number in lowest terms. b a B1 . There exists integers r and s with s > 0 A4 . Let r = and s = . a r d d and in lowest terms such that is A5 . r and s are integers, by a s b r previous result. equivalent to . s A6 . s > 0 because b > 0 and d > 0. B2 . There exists integers r and s with s > 0 A7 . gcd (r; s) = 1 by a previous result and gcd (r; s) = 1 such that as = br. b a b = rb. = A8 . as = a Switch to forward. d d Done. The following theorem requires advanced techniques; we omit the proof. Theorem. If a; b; c are positive integers, gcd (a; b) = 1 and a divides bc, then a divides c. Example: (Not su¢ cient for a proof) a = 4, b = 3, c = 16. Then bc = 48, so we have gcd (a; b) = 1, and a divides bc. The theorem tells us that a divides c; 4 divides 16 is true, so it works in this example. New example: a = 7; b = 13, c = 21. then bc = 273. We have gcd (a; b) = 1. Also 7 divides 273. The conclusion is that 7 divides 21, which is true. a a Theorem. If a; b; c; d are positive integers, is in lowest terms and is equivalent b b c to , then there is an integer k such that ak = c and bk = d. d Proof: 2 a A: a; b; c; d are positive integers, is in b c a lowest terms and is equivalent to b d A1 . There are positive integers m and n, with m m in lowest terms, such that a =c n n m = d. (Previous results.) and b n switch to backward A3 . am = cn and bm = dn A4 . n divides am and n divides bm A5 . n divides a and n divides b (Since gcd (m; n) = 1 and using a previous theorem.) A6 . n is a common divisor of a and b. a A7 . gcd (a; b) = 1, since is in lowest terms. b A8 . n = 1. Done. B: there is an integer k such that ak = c and bk = d. B2 . n = 1 Switch to forward We now introduce a new technique: uniqueness. Use this technique to show some object is unique, or that there is only one object, or similar phrasing. The uniqueness method has two parts. Part 1. Show that the object exists. The A is the same as usual. The B is the same as usual except the the word “unique”(or similar expression) is omitted. Part 2. Show that there is only one such object. This can be done in one of two ways: (a) Assume there are two such objects, and show that they are equal. The A would include the original A, along with the original B, except that the existence of two objects is asserted. The B is that the objects are equal. (b) Assume there are two such objects and reach a contradiction. The A is the original A, the original B for two such objects that are not equal. There is no B. Theorem: If r and s are rational numbers and r 6= 0, then there is a unique rational number x such that rx = s. Proof: Part 1: Existence. B: there is a rational number x A: r and s are rational numbers and r 6= 0 s A1 . Let x = , we know that x is a rational such that rx = s. r number from a previous result. Done with Part 1. Part 2. Uniqueness, …rst method. 3 A: r and s are rational numbers and r 6= 0, there are rational numbers x and y such that rx = s and ry = s. A1 . rx = ry, since they both equal s A2 . x = y Done. B: x = y We don’t have to do the second method because we have already completed the proof, but let’s see how it works: A: r and s are rational numbers and r 6= 0, B: (nothing) there are rational numbers x and y with x 6= y such that rx = s and ry = s. A1 . rx = ry, since they both equal s A2 . x = y C! We now complete the development on rational numbers in lowest terms. With the Theorem below, we have given a rigorous justi…cation for a process we have been using since elementary school. m is Theorem. If m and n are positive integers, then the rational number n equivalent to a unique rational number in lowest terms. Uniqueness method. Part 1: existence. We did this earlier in this module. Part 2. Uniqueness. m is B: a = c and b = d A: m and n are positive integers and the rational number n a c B3 . r = 1 equivalent to two rational numbers and in lowest terms. back to forward. b d Here a; b; c; d are positive integers. a c A1 . is equivalent to b d A2 . There is a positive integer r such that ar = c and br = d (result from earlier in the module, a use the fact that is in lowest terms.) b switch to backward. A4 . r is a common divisor of c and d. A5 . gcd (c; d) = 1. A6 . r = 1 because r is a positive integer and r gcd (c; d) Done. Questions. 4 1. Show that the rational numbers 12 20 , are equivalent. 18 30 4 5 and are not in lowest terms. Explain why. 8 2 Questions 3 and 4 refer to the proof of “If a; b; c; d are integers, b 6= 0; d 6= 0, a c and is equivalent to , then there is a rational number r such that ar = c b d and br = d.” 2. The rational numbers 3. How does step A1 follow from the hypothesis? 4. What remains to be proved after step A4 ? How does step A5 accomplish this? Questions 5 through 7 refer to the proof of “Theorem. If a and b are positive a integers, then the rational number is equivalent to a rational number in lowest b terms”. 5. Explain how we got step B1 . 6. What de…nition leads to step B2 . 7. Step A7 refers to a previous result. What result is this? 8. After stating “Theorem. If a; b; c are positive integers, gcd (a; b) = 1 and a divides bc, then a divides c” we gave two examples, and stated that they were not su¢ cient for a proof. Explain why. Questions 9 through 11 refer to the proof of “Theorem. If a; b; c; d are positive a a c integers, is in lowest terms and is equivalent to , then there is an integer b b d k such that ak = c and bk = d”. 9. We started with the forward method. How does step A1 follow from the hypothesis? 10. Explain step B2 . 11. Step A4 says “n divides am and n divides bm”. Justify this step. 12. The uniqueness method, part 2, says “Assume there are two such objects”. Explain how this was implemented in the proof of “Theorem: If r and s are rational numbers and r 6= 0, then there is a unique rational number x such that rx = s.” 13. In the proof of “Theorem. If m and n are positive integers, then the rational m number is equivalent to a unique rational number in lowest terms”, step B3 n tells us that it su¢ ces to show that r = 1. Explain why. 5 Problems In problems 1 –3, write three rational numbers equivalent to the given rational number. 1. 5 2. 3. 4. 8 12 10 6 In problems 4 –6, write the given rational numbers in lowest terms: 21 49 5. 18 8 6. 120 15 7. 6 17 8. Prove: If a and b are nonzero integers and a divides b, then there exists a unique integer k such that b = ak. 9. Prove: If a is an odd integer, there exists a unique integer b such that a = 2b 7. 10. Prove: If n is an even integer, there exists a unique integer k such that n = 2k + 12. 11. Prove: If x and y are rational numbers, then there is a unique rational z such that z = 2x + 5y. 6 
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