Math 4: Principles of Analysis Fall 0 Homework Part B solutions. Let x, y, z R. Use the axioms of the real numbers to prove the following. a) If x + y = x + z then y = z. Solution. By Axiom a), there is a number x such that x + x) = 0. Adding x to the equation x+y = x+z gives x)+x+y) = x)+x+z). By Axiom, this becomes x)+x)+y = x)+x)+z. By Axiom, x)+x = x+ x) = 0, so we have 0 + y = 0 + z, which then becomes y + 0 = z + 0 by Axiom, and then y = z by Axiom 4a). b) x) = x. Solution. By Axiom a), the number x) is the unique number such that x + x)) = 0. But x + x = x + x) b Axiom, and x + x) = 0 by Axiom a), so x + x = 0, so by uniqueness, x = x). c) If x 0, then x ) = x. Solution. By Axiom b), since x 0, there is a unique number x such that x x =. If x = 0, we would have x 0 =. But by Theorem.3.f) this would imply 0 =, which contradicts Axiom 4b). Thus x 0, so by Axiom b) there exists a number x ) such that x x ) =. Multiplying both sides of this by x gives by Axiom 4b). On the other hand, Thus x ) = x. x x x ) ) = x = x x x x ) ) = x x )) x ) Axiom = x ) Axiom b) = x ) Axiom = x ) Axiom 4b). d) x + y) = x) + y) Solution. By Axiom a), x + y) is the unique number such that x + y) + x + y)) = 0. Thus, it suffices to show that x + y) + x) + y)) = 0. To
prove this, we write x + y) + x) + y)) = x + y + x) + y))) Axiom = x + y + y) + x))) Axiom = x + y + y)) + x)) Axiom = x + 0 + x)) Axiom a) = x + x) + 0) Axiom = x + x) Axiom 4a) = 0 Axiom a) e) x y) = x) y. Solution. By Axiom a), x y) is the unique number such that x y+ x y)) = 0. Thus we just need to show that x y + x) y = 0. But x y + x) y = y x + y x) Axiom = y x + x)) Axiom 3 = y 0 Axiom a) = 0 y Axiom = 0 since we proved in class that 0 y = 0 for any y R. f) If x y = 0 then either x = 0 or y = 0. Solution. Suppose x y = 0. If y = 0 then we are done. Otherwise y 0, and Axiom b) implies that there is a real number y such that y y =. Thus x = x = x y y ) Axiom 4b) = x y) y Axiom = 0 y hypothesis = 0 again since 0 x = 0 for any real number x. g) If x > 3 then x 3 + x > 33. Solution. First, since > 0, Axioms and 9 imply x = x > 3 =. Next, since 3 > 0, Axiom 7 implies x > 0. Thus, Axiom 9 implies x > 3 x. But Axiom implies 3 x = x 3 and Axiom 9 implies x 3 > 3 3 = 9. Hence Axiom 7 implies x > 9. Axioms 9 and then imply x 3 > 9 x = x 9 > 3 9, so Axiom 7 implies x 3 > 7. Finally, Axiom 8 implies x 3 + x > 7 + = 33. h) If 0 < x <, then x < x. Solution. Since x > 0, Axiom 9 implies x = x > x x = x.
i) xy = x y Solution. If x 0 and y 0, then xy 0, so xy = xy = x y. If x < 0 and y < 0, then 0 = x + x) < 0 + x) = x, so x > 0 and likewise y > 0. Thus x) y) > 0 by Axiom 9. Thus x y = x) y) = x) y) > 0, so xy = xy and x y = x) y) = xy. If x 0 and y < 0, then x = x and y = y. Axiom 9 implies x y < 0, so xy = xy) = x y) = x y. The case x < 0 and y 0 is similar.. Use induction to show that k = Solution. Let S n denote the statement: nn + )n + ) k = In the base case n =, statement S takes the form for all n N. nn + )n + ). k = both sides of the equation equal, the statement S is true. + ) + ). Since Now suppose that S n is true for some integer n. We then need to prove that S n+ n+ must also be true. Now S n+ is the statement k n + )n + )n + 3) =. If we consider the left hand side of this equation, and use the fact that S n is true, we have n+ ) k = k + n + ) nn + )n + ) = + n + ) = n + nn + ) + n + )) = n + n + 7n + ) n + )n + )n + 3) =. Therefore the statement S n+ is true. Hence the implication S n = S n+ holds for all n, so by the principle of induction S n is true for all n. ) nn + ) 3. Use induction to show that k 3 = for all n N. ) nn + ) Solution. Let S n denote the statement: k 3 =. ) + ) In the base case n =, statement S takes the form k 3 =. Since both sides of the equation equal, the statement S is true. 3
Now suppose that S n is true for some integer n. We then wish to prove that n+ ) n + )n + ) S n+ must also be true. Now S n+ is the statement k 3 =. If we consider the left hand side of this equation, and use the fact that S n is true, we have n+ ) k 3 = k 3 + n + ) 3 ) nn + ) = + n + ) 3 n + ) = n + 4n + ) ) 4 n + ) = n + ) 4 ) n + )n + ) =. Therefore the statement S n+ is true. Hence the implication S n = S n+ holds for all n, so by the principle of induction S n is true for all n. 4. Use induction to show that Solution. The statement S reads both sides are equal to. kk + ) = n n + for all n N. Now suppose S n is true for some n. Then n+ ) kk + ) = + kk + ) n + )n + ) = n n + + n + )n + ) nn + ) = n + )n + ) + n + )n + ) = n + n + n + )n + ) = n + n +, kk + ) =. This is clearly true since + since S n is true so S n+ is true. Therefore S n = S n+ for all n, and the principle of induction implies S n is true for all n.. Find an integer n 0 such that n < 3 n for all n n 0. Use induction to prove your assertion. 4
Solution. Let S n denote the statement n < 3 n. By trial and error, we see that n 3 n if n 0, but that = 0 < 7747 = 3. Thus S n is false if n 0 but S is true. We claim that S n is true for all n. The base case S is true, so now fix some n and suppose S n is true for this n. We want to show that S n+ is true, i.e. that n + ) < 3 n+. Since S n is true we have so multiplying both sides by 3 gives n < 3 n, 3n < 3 3 n = 3 n+. If we can show that n + ) < 3n, then by transitivity we would have n + ) < 3 n+ as desired. But n + ) < 3n n + ) < 3 + < 3. n n) Now since n we have + + = n Thus we have + n) < and therefore ) = 4883 0 < 3. n + ) < 3n < 3 n+ and therefore S n+ is true. Hence the implication S n = S n+ is true for all n so by the principle of induction S n is true for all n.. Let x = 0 and let x n+ = + 3 x n for n. a) Compute x n for n =, 3, 4,. Solution. x = + 3 x = + 3 0) = x 3 = + 3 x = + ) = 0 3 x 4 = + 3 x 3 = + 38 0) = 3 3 x = + 3 x 4 = + 30 38/3) = 3 9 b) Prove by induction that x n < 8 for all n N. Solution. Let S n be the statement x n < 8. Since x = 0 < 8, S is true. Now suppose S n is true for some n. Then x n < 8. Multiplying this inequality by gives x 3 3 n <. Adding to both sides then gives + x 3 n < 8. By definition x n+ = + x 3 n, so x n+ < 8, and thus S n+ is true. Hence S n = S n+ for all n, so by induction S n is true for all n.
c) Prove by induction that x n+ > x n for all n N. Solution. Let S n be the statement x n+ > x n. Since x = > 0 = x, S is true. Now suppose S n is true for some n. Then x n+ > x n. Multiplying this inequality by gives x 3 3 n+ > x 3 n. Adding to both sides then gives + x 3 n+ > + x 3 n. By definition x n+ = + x 3 n and x n+ = + x 3 n+, so x n+ > x n+, and thus S n+ is true. Hence S n = S n+ for all n, so by induction S n is true for all n. 7. Let A = { n+3 : n N}. Find luba) and glba), and prove your assertions. Solution. For any n, we have n + 3 4, so, which implies, n+3 4 n+3 4 and thus 3. Thus 3 is a lower bound for A. But 3 A take n = ), so any n+3 4 4 4 number larger than 3 is not a lower bound for A. Thus glba) = 3. 4 4 Next, since < for all n N, is an upper bound for A. Now suppose x <. n+3 Then x > 0. By the Archimedian Property, there exists some n N such that n > 3. Thus n + 3 >, and since x > 0 this implies x)n + 3) >. x x Since n + 3 > 0, this then implies x >, and thus > x. But is n+3 n+3 n+3 an element of A, so x is not an upper bound for A. Therefore luba) =. 8. Suppose A and B are nonempty subsets of R such that x y for every x A and every y B. Show that luba) glbb). Solution. There are a couple of ways to prove this. First, given x A, since x y for every y B it follows that x is a lower bound for B and thus x glb B. Since this is true for every x A it follows that glb B is an upper bound for A and therefore glb B lub A. Alternately, one could do a proof by contradiction. Suppose lub A > glb B. Then lub A is not a lower bound for B so there exists some y B such that y < lub A. Then this y cannot be an upper bound for A, so there must exist some x A such that x > y. This contradicts the initial assumption that x y for every x A and every y B.