Math 25a Homework 2 Solutions part 1 Ivan Corwin and Alison Miller. 1 Injections, surjections, bijections (1) Prove that the composition of two injective functions is an injective function and that the composition of two surjective functions is a surjective function. Solution. Suppose f and g are injections. f ◦ g(x) = f ◦ g(y) ⇒ g(x) = g(y) ⇒ x = y, so f ◦ g is an injection. Suppose f and g are surjections. Consider c ∈ Img(g). ∃b s.t. g(b) = c and subsequently a s.t. f(a) = b. Then f ◦ g(a) = c, so f ◦ g is a surjection. (2) Let f : A → B be a function. Show that the following are equivalent. (a) There exists a function g : B → A such that g(f(x)) = x for all x ∈ X and f(g(y)) = y for all y ∈ Y . (b) f is a bijection. Solution. ⇒: Suppose f(x) = f(y). Then x = g(f(x)) = g(f(y)) = y, so f is injective. Take any y. f(g(y)) = y, so there ∃x s.t. f(x) = y. So f is surjective. Therefore, f is bijective. ⇐: Take g(y) = the element inf −1(y) for y ∈ B. Since f is a bijection, we know that f −1(y) exists and consists of exactly one element, so g is well-defined. Then g(f(x)) = x and f(g(y)) = y follows. (3) Let f : R → R be a function. We say that f is strictly increasing if f(x) < f(y) whenever x < y. (a) Show that if f is strictly increasing, it is an injection. Must it be a surjection? (b) Suppose f is a bijection and f(0) = 0 and f(1) = 1. Does it follow that f is strictly increasing? Solution. • Suppose f(x) = f(y) and x 6= y, then WLOG x > y. But by definition of strictly increasing, f(x) > f(y), a contradiction. So x = y, and f is an injection. It does not have to be a surjection, however, as exemplified by ( x − 1 x ≤ 0 f(x) = x x > 0 1 . • No. Consider ( 1/x x 6= 0 f(x) = 0 x = 0 (5) Give an explicit bijection between each of the following sets (see Rudin page 31 for the notation): (a) [1, 2] and [3, 7] (b) (0, 1) and (0, ∞) (c) [0, 1] and [0, 1) (d) R × R and R (Hint: see decimal expansions ... ) Solution. (a) Let X = [1, 2], Y = [3, 7]. Consider f : X → Y , f(x) = 4x − 1. (b) (0, 1) and (0, ∞) Let X = (0, 1), Y = (0, ∞). Consider f : X → Y , f(x) = 2 sin(πx)/(1 − cos(πx)). In fact, f takes every point with positive abscissa on the unit circle centered around (0, 1), and maps it to the intersection of the line through the point and (0, 2) and the x-axis. (c) [0, 1] and [0, 1) Let X = [0, 1], Y = [0, 1). Take a countable subset {xi} of X with x1 = 1 (letting xi = 1/i, for example). Let f take xi to xi+1, and x to x ∀x∈ / {xi}. (d) R × R and R We know that there is a bijection from (0, 1) to R by taking f(x) = tan((x − 1/2)π). It then suffices to supply a bijection from (0, 1) × (0, 1) to (0, 1). To do this, we first consider each element in (0, 1) as a decimal, using an infinite sequence of 0’s instead of 9’s when applicable. Then we divide x ∈ (0, 1), x = 0.x1x2x3 ... into pieces x = 0.X1X2X3 ..., where each piece starts from the end of the previous piece and ends when it encounters a non-9 digit (which exists by our construction of the decimals). Therefore, for each pair (x, y) ∈ (0, 1) × (0, 1), we can split x and y into 0.X1X2X3 ... and 0.Y1Y2Y3 .... Then construct z = 0.X1Y1X2Y2 .... This is a bijection since it cannot end in repeating 9’s, and it is a reversible process. 2 Fields, rational and irrational numbers (1) Given x ∈ R with x > 0 and an integer k ≥ 2, define a0, a1,... recursively by setting a0 = bxc (this means the largest integer which is less than or equal to x) and an to be the largest integer such that a a a a + 1 + 2 + ··· + n ≤ x 0 k k2 kn (a) Show that 0 ≤ ai ≤ k − 1 for each i ≥ 1. a1 an (b) Let rn = a0 + k + ... kn . Show that sup{r0, r1,... } = x. a1 a2 (Note: The expression r = a0 + k + k2 + ... is called the base-k expansion of x ∈ R. If k = 10, this is the decimal expansion of x. The next part of the problem shows that the base-k expansion of x is almost unique.) (c) Show that if we have sequences a0, a1,... and b0, b1,... such that (I) 0 ≤ ai ≤ k − 1 and 0 ≤ bi ≤ k − 1 2 a1 a2 b1 b2 (II) a0 + k + k2 + ··· = b0 + k + k2 + ... (III) for each N > 0 there exists and n > N and m > N such that an 6= k − 1 and bm 6= k − 1 then ai = bi for all i. (Hint: This last condition just says that neither sequence ends with an infinite sequence of 2 1 (k − 1)’s. You may need to use the fact that if 0 ≤ |x| < 1, then 1 + x + x ··· = 1−x .) Solution. (a) Since a a a a + 1 + 2 + ... + n ≤ x 0 k k2 kn Taking an+1 = 0 gives a a a a + 1 + 2 + ... + n+1 ≤ x 0 k k2 kn+1 and since we are picking the largest possible an+1, an+1 ≥ 0. Suppose an+1 > k − 1. Then an+1 1 kn+1 ≥ kn . So for an we could have picked an+1 instead, since that gives an extra contribution 1 of kn . So an+1 ≤ k − 1. 0 0 (b) Clearly x ≥ rn∀n. Now, we just need to make sure there does not ∃x < x s.t. x ≥ rn∀n. Take 1 0 n large enough s.t. kn ≤ x − x . Therefore, adding 1 to an would still satisfy our inequality in the construction of an, which is a contradiction since we picked the largest possible an. (c) Consider a a t = n + n+1 + ... n kn kn+1 . An unattainable upper bound of of tn occurs when all of the ai equal k − 1, which gives k−1 1 1 1 kn (1 + k + k2 + ...) = kn−1 . We define {sn} similarly, with bn in place of an as they occur. ai bi We take the smallest i s.t. ai 6= bi. WLOG, ai < bi. Then ki + ti+1 = ki + si+1. However, bi ai 1 1 1 i − ≥ i and t − s < i+1−1 − 0 = i , a contradiction. So a = b ∀i. k ki k i+1 i+1 k k i i (2) Use the elementary order axioms to prove the following properities: (a) If a ≤ b and c ≤ d, then a + c ≤ b + d. (b) If x, y ≥ 0 and x2 > y2, then x > y. (c) R is not bounded above. Solution. (a) Since a ≤ b it follows that a + c ≤ b + c. Similarly since c ≤ d it follows that c + b ≤ d + b. Transitivity implies then that a + c ≤ b + c ≤ b + d and therefore a + c ≤ b + d. (b) Assume from the point of contradiction that x ≤ y. Therefore 0 ≤ x ≤ y. Then since y − x ≥ 0 and x ≥ 0 it follows that x(y − x) ≥ 0, or that xy ≥ x2. Similarly since x ≤ y and y ≥ 0, xy ≤ y2. Therefore x2 ≤ xy ≤ y2 which implies x2 ≤ y2, a contradiction. Thus x > y. (c) Assume from the point of contradiction that R is bounded from above. Therefore it has a supremum a ∈ R. R also contains a positive element (such as 1) and hence by ordering (first part of the question), a + 1 > a. Yet a + 1 ∈ R by closure under addition. Therefore a is not the supremum, a contradiction. 3 (3) (a) Prove that there is exactly one way to make Q into an ordered field. Hint: Assume that there is another order x ≺ y on Q satisfying the axioms for an ordered field and show that x ≺ y if and only if x < y. (b) Let p be prime. Show that the field Z/pZ cannot be made into an ordered field. Show that C cannot be made into an ordered field. Solution. We know that 0 ≺ 1, from which a ≺ a + 1∀a ∈ Z. By transitivity a ≺ b iff a < b, so a ≺ 0 ⇔ a < 0 ⇔ 1/a ≺ 0 ⇔ 1/a < 0 (since (1/a)a = 1). This gives us b/a ≺ 0 ⇔ b/a < 0∀{a, b}. So a/b ≺ c/d ⇔ (ad − bc)/(bd) ≺ 0 ⇔ (ad − bc)/(bd) < 0 ⇔ a/b < c/d. Z/pZ: Suppose it can be made into an ordered field. Then 0 < 1. Adding 1 to both sides gives 1 < 2. Continuing on, 0 < 1 < 2 . p = 0, a contradiction. C: Suppose it can be made into an ordered field. Then i > 0 or i < 0. In either case, i2 > 0. But i2 = −1 < 0, a contradiction. √ √ (4) (a) Show that 2 + 3 is irrational. (b) If a and b is irrational, must a + b be irrational? Solution. (a) √ √ √ 2 + 3 ∈ Q ⇒ 2 + 3 + 2 6 ∈ Q √ ⇔ 5 + 2 6 ∈ Q √ ⇔ 6 ∈ Q √ Suppose 6 = m/n,(m, n) = 1. Then 6n2 = m2, so 2|m.