Homework 1 Solutions

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1 MATH 171 Spring 2016 Problem 1 Homework 1 Solutions (If you find any errors, please send an to farana at stanford dot edu) Presenting your arguments in steps, using only axioms of an ordered field, (listed as Axioms 1-12 in the text, on pages 10-12) and stating which property is used at each step, prove the following: i) ( 1)a = a. Solution. By the definition of the additive inverse a (Axiom 5), we need to check that a + ( 1)a = 0 = ( 1)a + a. As addition is commutative (Axiom 3) it is enough to check that a + ( 1)a = 0. As multiplication distributes over addition (Axiom 11) we have a + ( 1)a = a(1 + ( 1)). By the definition of the additive inverse 1 (Axiom 5) we have 1+( 1) = 0. It follows that a+( 1)a = a0. It remains to show that a0 = 0. This is Theorem 3.4 of the textbook. As it is a good exercise, we also present a proof here. By definition of the additive identity (Axiom 4) we have 0 = In particular a0 = a(0 + 0). As multiplication distributes over addition (Axiom 11) we have a(0 + 0) = a0 + a0. It follows that a0 = a0+a0. Adding the multiplicative inverse (a0) on both sides of the previous equality we get a0+( (a0)) = (a0+a0)+( (a0)). As additions is associative (Axiom 2) we can group the terms in the right hand side of the previous equality to get a0+( (a0)) = a0+(a0+( (a0))). Now by the definition of the additive inverse (a0) (Axiom 5) we have a0+( (a0)) = 0. Using this in the equality a0 + ( (a0)) = a0 + (a0 + ( (a0))) we get 0 = a But a0 + 0 = a0 by the definition of additive identity (Axiom 4). We then see that a0 = 0. Recall that we had a + ( 1)a = a0. Using the fact a0 = 0 just proved we get a + ( 1)a = 0, which is precisely what we wanted to show. ii) a > 0 1/a > 0. Solution. Notice that a > 0 implies a 0 (Axiom 12). It follows that a has a multiplicative inverse 1/a (Axiom 10). Exactly one of the following must hold: a < 0, a = 0, or a > 0 (Axiom 12). We must discard the first two cases. Suppose by contradiction that 1/a < 0, i.e. (1/a) > 0. Axiom 12 ensures that ( (1/a))a > 0. Now notice that ( (1/a))a = (( 1)(1/a))a by Part i). By associativity of multiplication (Axiom 7) we have (( 1)(1/a))a = ( 1)((1/a)a). Notice that (1/a)a = 1 by the definition of the multiplicative inverse 1/a (Axiom 12). It follows that ( 1)((1/a)a) = ( 1)1. By the definition of the multiplicative identity 1 (Axiom 9) we have ( 1)1 = 1. It follows from these calculations that ( (1/a))a = 1. We would then have 1 > 0. This contradicts the fact 1 > 0 (Theorem 4.2) and Axiom 12. Suppose now that 1/a = 0. Multiplying this equality by a we get (1/a)a = 0a. Notice that (1/a)a = 1 by the definition of the multiplicative inverse 1/a (Axiom 12). By the same arguments as in Part i) (or Theorem 3.4 of the textbook) we see that 0a = 0. We would 1

2 then have 1 = 0, contradicting the definition of the multiplicative identity 1 (Axiom 9). It follows that 1/a > 0, which is what we wanted to prove. iii) The complex numbers, C, do not form an ordered field; that is, there is no subset P C such that Axiom 12 holds. Solution. Suppose by contradiction there exists a subset P C such that Axiom 12 holds. Let i = 1 C. Then Axiom 12 ensures i > 0, i = 0, or i < 0. We show none of these can happen. Suppose first that i > 0. Then Axiom 2 ensures ii > 0. But ii = 1. We would then get 1 > 0. This a contradiction with the fact 1 > 0 (Theorem 4.2) and Axiom 12. Suppose now that i = 0. Then, using the same arguments as in Part i), we have 1 = ii = i0 = 0. It follows that 1 = 0. Adding 1 to both sides of the equality we get ( 1)+1 = 1. But ( 1)+1 = 0 by the definition of the additive inverse 1 (Axiom 5). It follows that 1 = 0, contradicting the definition of the multiplicative identity 1 (Axiom 9). Suppose now that i < 0, i.e. i > 0. Then Axiom 2 ensures ( i)( i) > 0. Using Part i) we see that ( i)( i) = (( 1)i)( i). As multiplication is commutative (Axiom 8) we deduce (( 1)i)( i) = (i( 1))( i). As multiplication is associative (Axiom 7) we have (i( 1))( i) = i(( 1)( i)). Using Part i) again we see that ( 1)( i) = ( i). We claim ( i) = i. By the definition of additive inverses (Axiom 5) we need tot check that ( i) + i = 0 = i + ( i). But this is true precisely because of the definition of the additive invere i (Axiom 5). The previous calculations show that ( i)( i) = i 2. We know that i 2 = 1. We would then have 1 > 0. This contradicts the fact 1 > 0 (Theorem 4.2) and Axiom 12. We conlcude that C is not an ordered field. Problem 2 Prove that every positive real number a > 0 has a positive square root. (Hint: Let S = {x R: x > 0 and x 2 < a}, and begin by showing that S is non empty and bounded above.) Proof. As the hint suggests we consider the set S = {x R: x > 0 and x 2 < a}. We claim S. Indeed, let r = min{1, a}. Consider any x R such that 0 < x < r. Then x > 0 and x 2 < 1 a = a, so that x S. We claim that S is bounded above by t = max{1, a}. Indeed, let y S and suppose by contradiction that y > t. Then y 2 > t 2 > 1a = a, contradicting the fact that y S. The least upper bound axiom (Axiom 13) ensures the existence of a least upper bound s of the set S. We claim that s 2 = a. Notice that s r > 0 because x S for all x R such that 0 < x < r. Notice also that s t because t is an upper bound of S. We show that s 2 a. Suppose by contradiction that s 2 < a. Consider an arbitrary small number 0 < ε < 1. Notice that (s + ε) 2 = s 2 + 2sε + ε 2 s 2 + 2tε + ε = s 2 + (2t + 1)ε. As s 2 < a we can find 0 < ε < 1 small enough so that s 2 + (2t + 1)ε < a. It follows that s + ε S. As s + ε > s this contradicts the fact that s is an upper bound of S. This proves that s 2 a. We now show that s 2 a. Suppose by contradiction that s 2 > a. Consider an arbitrary small number 0 < ε < 1. Notice that (s ε) 2 = s 2 2sε + ε 2 > s 2 2tε. As s 2 < a we can find 0 < ε < 1 small enough so that s 2 2tε > a. In particular (s ε) 2 > a. We claim that (s ɛ) is an upper bound of S. Suppose 2

3 by contradiction there existed some x S such that (s ɛ) < x. It follows that (s ɛ) 2 < x 2 < a, which contradicts the fact (s ε) 2 > a. We then see that s ε is an upper bound of S. As s ε < s this contradicts the fact that s is the least upper bound of S. This proves that s 2 a. We conclude that s 2 = a. Problem 3 Let X be a set of real numbers with least upper bound a. Prove that if ε > 0, there exists x X such that a ε < x a. Solution. As a is a least upper bound of X, it is in particular an upper bound of X, i.e. x a for all x X. Suppose by contradiction that the property we are trying to prove does not hold. Then for some ε > 0 and for all x X we cannot have a ε < x a. As we have x a for all x X this implies that a ε x for all x X. But then a ɛ is by definition an upper bound of X. Notice also that a ɛ < a because ɛ > 0. This contradicts the fact that a is the least upper bound of X. We conclude that the desired property holds. Problem 4 Let X and Y be sets of real numbers with least upper bounds a and b, respectively. Prove that a + b is the least upper bound of the set X + Y = {x + y x X, y Y } Solution. We first show that a + b is an upper bound of X + Y. Let x X and y Y be arbitrary, so that x + y is an arbitrary element of X + Y. As a is a least upper bound of X it is in particular an upper bound of X. Then we must have x a. By the same argument y b. It follows that x + y a + b. We see then that a + b is an upper bound of X + Y. We now show that a + b is the least upper bound of X + Y. Let c R be an arbitrary upper bound of X + Y, i.e. x + y c for all x X and y Y. We show that a + b c. Fix for now some arbitrary y Y. Notice that x c y for all x X, i.e. c y is an upper bound of X. As a is the least upper bound of X we then have a c y. As y Y was arbitrary this holds for all y Y. Then we have y c a for all y Y, i.e. c a is an upper bound of Y. As b is the least upper bound of Y we get b c a. It follows that a+b c, which is what we wanted to prove. Problem 5 Prove that the set of real numbers as defined by axioms 1 to 13 is unique in the following sense: If R is a set which satisfies axioms 1 to 13, there exists a one to one function f from R onto R 3

4 which satisfies f(x + y) = f(x) + f(y) f(xy) = f(x)f(y) f(x) < f(y) if and only if x < y for all x, y R for all x, y R for x, y R Proof. Following Definition 6.3 we see that R has a set of positive integers P. By the same arguments as in Section 6 it makes sense to denote P := {1, 2,... } with elements presented in increasing order. We define { } p Q := q P, p P P {0} q We consider the map f : Q Q given by f (p/q) = p /q. It is straightforward to check that this map is well defined and bijective. We now extend this map in the following way: let f : R R be the map given by f(x) := lub{f(r) r < x, r Q}. We claim that f has the desired properties. Surjectiveness and injectiveness are easy to check using the fact that elements of R and R can be approximated by elements of Q and Q respectively as in Theorem 7.8. We focus on checking the arithmetic properties of f. It is straightforward to check that such properties hold for elements x, y Q. We extend such properties to elements x, y R. Let x, y R. We show that f(x + y) = f(x) + f(y). Using Problem 4 we see that f(x) + f(y) = lub{f(r) r < x, r Q} + lub{f(r ) r < y, r Q} = lub{f(r) + f(r ) r < x, r < y, r, r Q} = lub{f(r + r ) r < x, r < y, r, r Q} = lub{f(r) r < x + y, r Q} = f(x + y). The same argument with multiplication instead of addition and with special care regarding signs shows that f(xy) = f(x)f(y). Suppose now that x < y. By Theorem 7.8 we can find a rational numbers r 1, r 2 Q such that x < r 1 < r 2 < y. Notice then that f(x) = lub{f(r) r < x, r Q} f(r 1 ) < f(r 2 ) lub{f(r) r < y, r Q} = f(y). It follows that f(x) < f(y). Suppose now that f(x) < f(y). Then theorem 7.8 applied to R and Q lets us find r = p /q Q such that f(x) < r < f(y). We claim that r := p/q Q is such that x < r < y. Suppose by contradiction that r x. Then, by what was proved just before, we have r = f(r) f(x), which contradicts the fact f(x) < r. The same argument shows that r < f(y). We conclude that x < y. This completes the proof of the desired properties. Problem 6 Prove that if {a n } is an increasing sequence which is not bounded, then lim n a n =. Solution. Recall that a sequence {b n } is bounded if there exists some M R such that b n M for all n N. Then, {a n } being not bounded, for all M N there exists an n M N such that a nm > M. But a n is increasing, so we must have a k > M for all k n M. This is exact the defintion of lim n a n =. 4

5 Problem 7 Prove the nested interval theorem: If {[a n, b n ]} is a sequence of closed intervals such that [a n, b n ] [a n+1, b n+1 ] for n = 1, 2,... then [a n, b n ] is nonvoid. Find a necessary and sufficient condition that n=1 [a n, b n ] contains exactly one point. Solution. Notice that the sequence {a n } is increasing, while the sequence {b n } is decreasing. Notice also that the sequence {a n } is bounded above by b 1, while the sequence {b n } is bounded below by a 1. Then both sequences {a n } and {b n } converge (Theorem 16.2), say a n a and b n b. Notice also that a b, as is seen by taking limits in the inequality a n b n. It follows that [a, b] = [a n, b n ]. We conclude that [a n, b n ] is nonvoid. We claim that, in the setting above, n=1 [a n, b n ] contains exactly one point if and only if lim n b n a n = 0. Indeed, notice that lim n b n a n = 0 if and only if lim n b n lim n a n = 0 if and only if a b = 0 if and only if a = b if and only if [a, b] = {a} if and only if n=1 [a n, b n ] = {a}. Problem 8 Let 0 α < 1, and let f be a function from R into R which satisfies f(x) f(y) α x y for all a, y R Let a 1 R, and let a n+1 = f(a n ) for n = 1, 2,.... Prove that {a n } is a Cauchy sequence. Solution. We first explore a bit before choosing specific ε s. Notice that for any k N we have a k+1 a k = f k (a 1 ) f k 1 (a 1 ) α f k 1 (a 1 ) f k 2 (a 1 ) α k 1 f(a 1 ) (a 1 ). Consider arbitrary n, m N with m > n. Using the triangle inequality, the previous fact, and the fact that the geometric series of 0 α < 1 converges, we see that a m a n = m 1 k=n a k+1 a k m 1 k=n a k+1 a k m 1 k=n αk 1 f(a 1 ) (a 1 ) = f(a 1 ) (a 1 ) ( m 1 k=n αk 1 ) f(a 1 ) (a 1 ) ( k=n αk 1 ) = f(a 1 ) (a 1 ) α n 1 /(1 α). Consider the constant C = f(a 1 ) (a 1 ) /(1 α). Then we have a m a n Cα n 1 for any n, m N. As 0 α < 1 we know that α n 1 0 when n. But then, given ε > 0, we can always find N N such that for all n N we have α n 1 ε/c. It follows that for all n, m N we have a m a n ε. This shows that the sequence {a n } is a Cauchy sequence. Problem 9 Consider the set of rational numbers. Call two Cauchy sequences of rationals {a n } and {b n } equivalent, denoted {a n } {b n }, if lim n a n b n = 0. 5

6 a) Show that is an equivalence relation (symmetric, reflexive, and transitive). Solution. For any Cauchy sequence of rational numbers {a n } we have {a n } {a n }, because lim n a n a n = lim n 0 = 0. Let {a n } and {b n } be Cauchy sequences of rational numbers such that {a n } {b n }, i.e. lim n a n b n = 0. Then lim n b n a n = lim n a n b n = 0, so that {b n } {a n }. Now let {a n }, {b n }, and {c n } be Cauchy sequences of rational numbers such that {a n } {b n } and {b n } {c n }, i.e. lim n a n b n = 0 and lim n b n c n = 0. Notice then that lim n a n c n = lim n a n b n +b n c n lim n ( a n b n + b n c n ) = lim n a n b n + lim n b n c n = = 0, so that lim n a n c n = 0. It follows that {a n } {c n }. We conclude that is an equivalence relation. b) Define R to be the set of equivalence classes of Cauchy sequences of rationals. If x, y R, define x+y and xy, showing that you definition is independent of choices within an equivalence class. Solution. Let x = [{x n }] and y = [{y n }] be elements of R. We define x + y to be the equivalence class of the Cauchy sequence of rational numbers {x n + y n }, i.e. x + y = [{x n + y n }]. Notice that {x n + y n } is indeed a Cauchy sequence. Given ε > 0 let N 1 N be such that x m x n ε/2 for all n, m N 1 and N 2 N be such that y m y n ε/2 for all n, m N 2. Consider N = max{n 1, N 2 } N. Notice then that for all n, m N we have x m + y m (x n + y n ) = x m x n + y m y n = x m x n + y m y n ε/2 + ε/2 = ε. As ε > 0 was arbitrary, it follows that {x n + y n } is a Cauchy sequence. Notice that the equivalence class x + y = [{x n + y n }] is indepenendent of the choice of representatives {x n } of x and {y n } of y. Indeed, let {x n } {x n} and {y n } {y n} be arbitrary Cauchy sequences of rational numbers. We need to show that {x n + y n } {x n + y n}. For this notice that lim n x n y n (x n + y n) = lim n x n x n + y n y n lim n x n x n + y n y n = lim n x n x n + lim n y n y n = = 0. It follows that {x n + y n } {x n + y n}, and so x + y is independent of the choices of representatives within each equivalence class. Let x = [{x n }] and y = [{y n }] be elements of R. We define xy to be the equivalence class of the Cauchy sequence of rational numbers {x n y n }, i.e. xy = [{x n y n }]. Notice that {x n y n } is indeed a Cauchy sequence. As {x n } and {y n } are Cauchy sequences they are in particular bounded. Let M 1 N be such that x n M 1 for all n N and M 2 N be such that y n M 2 for all n N. Given ε > 0 let N 1 N be such that x m x n ε/2m 2 for all n, m N 1 and N 2 N be such that y m y n ε/2m 1 for all n, m N 2. Consider N = max{n 1, N 2 } N. Notice then that for all n, m N we have x m y m x n y n = x m y m x m y n + x m y n x n y n x m y m x m y n + x m y n x n y n = x m y m y n + y n x m x n M 1 ε/2m 1 + M 2 ε/2m 2 = ε. As ε > 0 was arbitrary, it follows that {x n y n } is a Cauchy sequence. Notice that the equivalence class xy = [{x n y n }] is indepenendent of the choice of representatives {x n } of x and {y n } of y. Indeed, let {x n } {x n} and {y n } {y n} be arbitrary Cauchy sequences of rational numbers. We need to show that {x n y n } {x ny n}. For this notice that lim n x n y n (x ny n) = lim n x n y n x n y n + x n y n x ny n lim n x n y n x n y n + x n y n x ny n = lim n x n y n y n + y n x n x n = lim n x n y n y n + lim n y n x n x n = = 0, where we used the fact that the sequences {x n } and {y n} are bounded (because they are Cauchy). It follows that {x n y n } {x ny n}, and so xy is independent of the choices of representatives within each 6

7 equivalence class. Problem 10 Let R be defined as in the previous problem. a) Show that R satisfies field Axioms 5 and 10: the existence of additive and multiplicative inverses. (It satisfies the other field axioms but you do not have to prove that). Solution. It is straightforward to check that the additive identity of R is given by the class of the constant sequence with value 0. We denote such class by 0. Let [{x n }] be an arbitrary element of R. We claim that [{ x n }] is the additive inverse of [{x n }] in R. Notice that { x n } is a Cauchy sequence of rational numbers so that [{ x n }] is indeed an element of R. From the definition of the operation + in R we see that [{x n }] + [{ x n }] = [{x n x n }] = [{0}] = 0 = [{0}] = [{ x n + x n }] = [{ x n }] + [{x n }]. It follows that [{ x n }] is indeed the additive inverse of [{x n }] in R. It is straightforward to check that the multiplicative identity of R is given by the class of the constant sequence with value 1. We denote such class by 1. Let [{x n }] be an arbitrary element of R 0. We claim that [{1/x n }] is the multiplicative inverse of [{x n }] in R. Here we consider the sequence {1/x n } as taking the value 1 whenever x n = 0. We first show that {1/x n } is a Cauchy sequence of rational numbers so that [{1/x n }] is indeed an element of R. To check this consider 1/x n 1/x m = x m x n / x n x m. Notice that {x n } does not converge to 0, else [{x n }] = 0. It follows that x nk = x nk 0 > δ for some δ > 0 and a sequence n k of positive integers. As {x n } is Cauchy there exists N 1 N such that x n x m < δ/2 for all n, m N 1. Considering any k N such that n k N we see that x n = x nk + x n x nk x nk x n x nk > δ δ/2 = δ/2 for all n N 1. Notice then that for all n, m N 1 we have 1/x n 1/x m = x m x n / x n x m < x n x m δ 2. As {x n } is Cauchy we can find N 2 N such that x n x m < ɛ/δ 2 for all n, m N 2. Consider N := max{n 1, N 2 }. Then 1/x n 1/x m < ɛ for all n, m N. It follows that {1/x n } is a Cauchy sequence of rational numbers. From the definition of the operation in R we see that [{x n }] [{1/x n }] = [{x n (1/x n )}] = [{1}] = 1 = [{1}] = [{(1/x n ) x n }] = [{1/x n }] [{x n }]. It follows that [{1/x n }] is indeed the multiplicative inverse of [{x n }] in R. b) Descibe a set P R of positive elements and show that it satisfies the order axiom, Axiom 12. Proof. Let P be the subset of R given by: P := {[{x n }]: {x n } is a Cauchy sequence of positive rational numbers bounded below from 0} By bounded below from 0 we mean x n > c for all n N and some c > 0. It is clear that x + y and xy are in P whenever x, y are in P. It remains to show that for any given x = [{x n }] R exactly one of the following three holds: x P, x = 0, or x P. It is easy to check that 7

8 x P if and only if there exists some N N and c > 0 such that x n > c for all n N. Analogously one checks that x P if and only if there exists some N N and c < 0 such that x n < c for all n N. It is clear then that the conditions x P, x = 0, or x P are mutually exclusive. It remains to check that any x R satisfies one of such conditions. Let x R be arbitrary. If x = 0 then we are done. Suppose that x 0. Notice that {x n } does not converge to 0, else [{x n }] = 0. It follows that x nk = x nk 0 > δ for some δ > 0 and a sequence n k of positive integers. As {x n } is Cauchy there exists an N 1 N such that x n x m < δ/2 for all n, m N. Considering any k N such that n k N we see that x n = x nk + x n x nk x nk x n x nk > δ δ/2 = δ/2 for all n N. In particular the x n with n N are either all positive or all negative ( x n x m < δ/2 for all n, m N). If all are positive we then have x n > δ/2 for all n N, i.e. x P. If all are negative we then have x n < δ/2 for all n N, i.e. x P. It follows that the set P satisfies the desired conditions. 8