Hints for Exercises from Section 5.2 –MATH 3000 Kawai [5.7] Prove that there is NO largest negative rational number. BWOC, suppose x is largest negative rational number. Let y = x=2: It is also negative and rational. If x < 0; then 1 1 < 1 x > 1 x; 2 ) 2 since we multiplied both sides by a negative number. Both sides are negative. x x < = y: ( ) 2 )( This contradicts that x is the largest negative rational number. [5.8] Prove that there is no smallest positive irrational number. Same as [5.7], but we need to show that if x is irrational, then x=2 is also irrational. EASY. Contrapositive! x If Q; then x Q: 2 2 2 We had a lemma that covers this. [5.9] BWOC, suppose 200 can be written as odd + even + even: ( ) )( [5.10] If a and b are odd, then 4 - a2 + b2 : Here’swhy indirect proof is so awesome. BWOC, we start with 4 a2 + b2 a2 + b2 = 4k: j ) If a and b are odd, then we use the usual representatives: a = 2k1 + 1; b = 2k2 + 1: This gives us: 2 2 (2k1 + 1) + (2k2 + 1) = 4k 2 2 4k1 + 4k1 + 4k2 + 4k2 + 2 = 4k Since this is a derived equality, we can manipulate both sides! 2 2 2 2 2 = 4k 4k 4k1 4k 4k2 = 4 k k k1 k k2 1 2 1 2 2 2 By closure, let m = k k k1 k k2; and this gives us: 1 2 2 = 4m 4 2: ( ) ) j )( [5.11] This should make sense. If a; b Z+ (it doesn’tmake sense for either of them to be negative or zero), and a 2; 2 then a - b or a - (b + 1) : Clearly, we will contradict a dividing BOTH of them. Hint: (( P ) ( Q)) P Q: _ , ^ 1 [5.13] Prove that the product of an irrational number and a nonzero rational number is irrational. Let our original irrational number be x and our original nonzero rational number be s: BWOC, suppose their product is rational, t: xs = t Divide both sides by s: t x = s The right side is rational divided by rational, and this must be rational. The left side is irrational. ( ) )( [5.19] Let S = p + qp2; p; q Q and T = r + sp3; r; s Q : 2 2 Clearly, if we …x q = 0; then we have a subset of S; and that subset is all of Q: The same happens when we …x s = 0 in T: Prove that S T = Q: \ Tricky: BWOC, suppose S T = Q: Thus, there must be an element in both S and T which is not rational. \ 6 [5.20] Prove that if x; y R+; then px + y = px + py: 2 6 BWOC, suppose 2 2 px + y = px + py px + y = px + py ) x + y = x + 2pxy + y 2pxy = 0 pxy = 0 xy = 0: ) ) ) What did we contradict? 2.