Direct Proof Rational Numbers

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Direct Proof Rational Numbers Lecture 14 Section 4.2 Robb T. Koether Hampden-Sydney College Thu, Feb 7, 2013 Robb T. Koether (Hampden-Sydney College) Direct Proof Rational Numbers Thu, Feb 7, 2013 1 / 22

1 Rational Numbers 2 Intersecting Lines 3 A More Challenging Problem 4 Assignment Robb T. Koether (Hampden-Sydney College) Direct Proof Rational Numbers Thu, Feb 7, 2013 2 / 22

Outline 1 Rational Numbers 2 Intersecting Lines 3 A More Challenging Problem 4 Assignment Robb T. Koether (Hampden-Sydney College) Direct Proof Rational Numbers Thu, Feb 7, 2013 3 / 22

Rational Numbers Definition (Rational Number) A rational number is a real number that can be represented as the quotient of two integers. That is, a real number r is rational if a, b Z, r = a b. Robb T. Koether (Hampden-Sydney College) Direct Proof Rational Numbers Thu, Feb 7, 2013 4 / 22

Properties of Rational Numbers Theorem The sum of two rational numbers is a rational number. Proof. Let r and s be rational numbers. Then there exist integers a, b, c, and d such that r = a b and s = c d. Then r + s = a b + c d = ad bd + bc bd ad + bc =. bd Robb T. Koether (Hampden-Sydney College) Direct Proof Rational Numbers Thu, Feb 7, 2013 5 / 22

Properties of Rational Numbers Proof. ad + bc and bd are integers. Furthermore, bd 0 because b 0 and d 0. Therefore, r + s is a rational number. Robb T. Koether (Hampden-Sydney College) Direct Proof Rational Numbers Thu, Feb 7, 2013 6 / 22

Properties of Rational Numbers Theorem Between any two distinct rational numbers there is a rational number. Let r and s be distinct rational numbers. We may assume that r < s. Let r = a b and s = c d Let for some integers a, b, c, and d. t = r + s 2 ad + bc = 2bd. ad + bc and 2bd are integers and 2bd 0, so t is a rational number. Robb T. Koether (Hampden-Sydney College) Direct Proof Rational Numbers Thu, Feb 7, 2013 7 / 22

Properties of Rational Numbers Proof. We claim that r < t < s. t = r + s 2 = 2r + (s r) 2 = r + s r 2 > r. Similarly, we can show that t < s. Therefore, t is between r and s. Robb T. Koether (Hampden-Sydney College) Direct Proof Rational Numbers Thu, Feb 7, 2013 8 / 22

Properties of Rational Numbers Theorem Let f (x) = c n x n + c n 1 x n 1 + + c 2 x 2 + c 1 x + c 0 be a polynomial of degreen n for some positive integer n, where c 0, c 1,..., c n be integers, and let r be a rational number. Then f (r) is a rational number. Proof. Let r = a b, where a and b are integers. Then ( a ) n ( a ) 2 ( a ) f (r) = c n + + c2 + c1 + c 0 b b b ( ) ( ) a n a 2 ( a ) = c n b n + + c 2 + c 1 + c 0 b b 2 = c na n + + c 2 a 2 b n 2 + c 1 ab n 1 + c 0 b n b n. Robb T. Koether (Hampden-Sydney College) Direct Proof Rational Numbers Thu, Feb 7, 2013 9 / 22

Properties of Rational Numbers Proof. Clearly, c n a n + + c 2 a 2 b n 2 + c 1 ab n 1 + c 0 b n and b n are integers and b n 0, so f (r) is a rational number. Robb T. Koether (Hampden-Sydney College) Direct Proof Rational Numbers Thu, Feb 7, 2013 10 / 22

Outline 1 Rational Numbers 2 Intersecting Lines 3 A More Challenging Problem 4 Assignment Robb T. Koether (Hampden-Sydney College) Direct Proof Rational Numbers Thu, Feb 7, 2013 11 / 22

Intersecting Lines The following equations describe lines in the plane, provided not both a and b are 0 and not both c and d are 0. ax + by = e cx + dy = f. If a, b, c, d, e, f Z, then must the lines point of intersection have rational coordinates? Robb T. Koether (Hampden-Sydney College) Direct Proof Rational Numbers Thu, Feb 7, 2013 12 / 22

Solve the System Solve the equations simultaneously. adx + bdy = de, bcx + bdy = bf. Then (ad bc)x = de bf. So x = What is wrong with that proof? de bf ad bc. Robb T. Koether (Hampden-Sydney College) Direct Proof Rational Numbers Thu, Feb 7, 2013 13 / 22

Outline 1 Rational Numbers 2 Intersecting Lines 3 A More Challenging Problem 4 Assignment Robb T. Koether (Hampden-Sydney College) Direct Proof Rational Numbers Thu, Feb 7, 2013 14 / 22

A More Challenging Problem Suppose that a, b, and c are integers and that there are nonzero real numbers x, y, and z satisfying xy x + y = a, xz x + z Must x, y, and z be rational? = b, and yz y + z = c. Robb T. Koether (Hampden-Sydney College) Direct Proof Rational Numbers Thu, Feb 7, 2013 15 / 22

Solve the Equations Take reciprocals: or x + y xy = 1 a, x + z xz = 1 b, and y + z yz = 1 c 1 y + 1 x = 1 a, 1 z + 1 x = 1 b, and 1 z + 1 y = 1 c. We can easily solve these for 1 x, 1 y, and 1 z. Robb T. Koether (Hampden-Sydney College) Direct Proof Rational Numbers Thu, Feb 7, 2013 16 / 22

Solve the Equations We get 1 x = 1 ( 1 2 a + 1 b 1 c 1 y = 1 ( 1 2 a 1 b + 1 c 1 z = 1 2 ), ), ( 1 a + 1 b + 1 c ). Robb T. Koether (Hampden-Sydney College) Direct Proof Rational Numbers Thu, Feb 7, 2013 17 / 22

Solve the Equations Thus, Check the answer. 2abc x = ab + ac + bc, 2abc y = ab ac + bc, 2abc z = ab + ac bc. Must the solutions be rational numbers? Robb T. Koether (Hampden-Sydney College) Direct Proof Rational Numbers Thu, Feb 7, 2013 18 / 22

Verify the Solution What if ab + ac + bc = 0? Then ab = ac + bc. Substitute into the formula for y and get 2abc y = ab ac + bc 2abc = (ac + bc) ac + bc = 2abc 2bc = a. Robb T. Koether (Hampden-Sydney College) Direct Proof Rational Numbers Thu, Feb 7, 2013 19 / 22

Verify the Solution Substitute y = a into the first of the original equations: xy x + y = a xa x + a = a x x + a = 1 x = x + a 0 = a. But our assumption was that a 0. Therefore, ab + ac + bc 0 and x is rational. A similar argument shows that y and z are rational. Robb T. Koether (Hampden-Sydney College) Direct Proof Rational Numbers Thu, Feb 7, 2013 20 / 22

Outline 1 Rational Numbers 2 Intersecting Lines 3 A More Challenging Problem 4 Assignment Robb T. Koether (Hampden-Sydney College) Direct Proof Rational Numbers Thu, Feb 7, 2013 21 / 22

Assignment Assignment Read Section 4.2, pages 163-168. Exercises 5, 7, 8, 14, 18, 22, 28, 30, 33, page 168. Robb T. Koether (Hampden-Sydney College) Direct Proof Rational Numbers Thu, Feb 7, 2013 22 / 22