MATH 201 Solutions: TEST 3-A (in class)

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1 MATH 201 Solutions: TEST 3-A (in class) (revised) God created infinity, and man, unable to understand infinity, had to invent finite sets. - Gian Carlo Rota Part I [5 pts each] 1. Let X be a set. Define P(X), the power set of X. P(X) is the set of all subsets of X. 2. Let F: X Y be a mapping. Define: F is injective. F: X Y is injective if: a, b X F(a) = F(b) a = b 3. Let G: X Y be a mapping. Define: G is surjective. G: X Y is surjective if: c Y a X G(a) = c n 4. Give the combinatorial definition of --- not the means of calculating this number or any other interpretation r using factorials. n r is the number of ways that an (unordered) subset can be chosen from a set of n items. 5. Let S be a set. Define: S is countably infinite. S is countably infinite if there exists a bijection from the set of natural numbers to S. 6. Let R and S be sets. What does it mean to assert that R and S are of the same cardinality? Two sets, R and S, are of the same cardinality if there exists a bijection from R to S

2 2 7. If A = 9876, then P(A) = If A = 9876, then P(A) = Let X and Y be sets and let G: X Y be a bijection. Define G -1 : Y X Let y Y. Since G: X Y is a bijection, there exists one and only one element x X for which G(x) = y. We define G(y) = x. Part II [10 pts each] Answer any 11 of the following 14. You may solve more for extra credit. 1. Here is a partial proof of Cantor s theorem. Your job is to complete it. Theorem: For any set X, P(X) and X are not of the same cardinality. Proof: Let X be a set and P(X) be the power set of X. Strategy: proof by contradiction. Assume that bijection F: X P(X). Define D = {q X q F(q)} Now, since F is surjective, x X for which F(x ) = D. Next, suppose that x D. Then, by definition of D, x F(x ) = D. This is impossible. Thus x D. But, D = F(x ); so x F(x ). This means that x D. But this too is impossible. Hence our assumption that F is a bijection is false. 2. Using the Euclidean algorithm, find gcd(210, 45) 210 = 45* = 30* = 15*2 + 0

3 gcd = Using the extended Euclidian algorithm, find integers x and y such that 56x + 22y = gcd(56, 22). 56 = 22* = 12* = 10* = 5*2 0 Hence gcd(56, 22) = 2 So 2 = = 12 (22 12*1) = 2*12 22 = 2(56 2*22) 22 = 2*56 4*22 22 = 2*56 5*22 Hence 56(2) 22(-5) = 2 4. Odette believes that the set of all finite sequences using the alphabet {a, b, c} is countably infinite. Give proof or counterexample. (For example: aab, cabcab, ccccccccccccccc are all finite sequences.) Odette is correct: Begin by listing all strings of length 1: a, b, c Then all 9 strings of length 2: aa, ab, ac, ba, bb, bc, ca, cb, cc Then all 27 strings of length 3, etc. 5. Swann believes that the set of all infinite sequences using the alphabet {a, b, c} is countably infinite. Give proof or counterexample. (For example: abababab, abcabccccbbaabbbbb are infinite strings.) Swann is mistaken. Just use Cantor s diagonal argument on the set of infinite strings: First string: E 11, E 12, E 13, E 14, Second string: E 21, E 22, E 23, E 24, Third string: E 31, E 32, E 33, E 34, Etc. Define the string e * as follows:

4 4 a if E i i = b or c e * i = { c if E i i = a Now notice that e * i cannot appear in our infinite list of strings 6. Explain briefly, using only a combinatorial interpretation, why about combinations.) (Use complete sentences. Do not use any formulas. Do not use more general results Let S be a set of 1776 objects. Then there is a bijection between the set of all subsets of S of size 99 and the set of all subsets of S of size This is true since selecting a subset of size 99 corresponds to selecting a subset of size = It is not difficult to see that this correspondence is a bijection. 7. Explain briefly, using only a combinatorial interpretation, why results about combinations.) (Use complete sentences. Do not use any formulas. Do not use more general Let S be a set of 1789 items. Call one of them a *. Then S = (S ~ {a * }) {a * }. Note that (S ~ {a * }) has 1788 items. Two cases (mutually exclusive): Case I: Choose 25 items from S ~ {a * }. Next, if we include {a * }). then we must choose the remaining 24 items from S. Case II: Choose all 25 items from (S ~ {a * }) Hence Given a group of 13 married couples. The number of ways we can choose a subset of 5 individuals from this group which contains no married couple is 13 choose 5 couples: 5 13 From each of the 5 chosen couples, choose 1 of the 2 partners. Hence the number of ways = 2 5 5

5 5 9. Let A, B be non-empty finite sets. Suppose that A =1789 and B = Then (a) The number of surjections from A into B is zero. Why? The range of any mapping from A into B cannot contain more than 1789 elements. (b) The number of injections from A into B is P(2016, 1789). Why? Each element of A can choose one of the remaining elements of B. In particular, the first element of A can select from 1789 elements; the second element of A can choose from 1788 elements; etc. 10. Let Q be the set of all rational numbers. Define f: Q Q as follows: (a) Is f well-defined? Justify your answer. 3 x f ( x) 0 if if x 0 x 0 Yes. For any non-zero rational number, x = p/q, we have f(x) = 3/x = 3q/p which is also rational. And, if x = 0, f(0) = 0 is also rational. (b) Is f injective? Why? Yes. Let x and y be non-zero rational numbers. Now assume that f(x) = f(y). Then 3/x = 3/y, which implies that x = y. (c) Is f surjective? Why? Yes: Let y be any non-zero rational number. Then 1/(3y) is also rational and f(1/(3y)) = y. 11. For any real numbers, c and d, let us define the binary operation as follows: c d = c 2 + d 2 1 Give either a brief justification or counterexample for each of the following assertions: (a) The set of integers is closed under the operation. Yes. Integers are closed under taking squares as well as under addition and subtraction.

6 (b) The set of even integers is closed under the operation. 6 number. No. For a counter-example, let c = 2 amd d = 4. Then c d = c 2 + d 2 1 = = 19, an odd (c) The set of odd integers is closed under. Yes If c and d are odd, then c 2 and d 2 are also odd (proven earlier in class). Since the sum of two odd numbers is even, we have c 2 + d 2 is even. Now c 2 + d 2 1 is odd. (d) The set of positive integers is closed under. Yes. If c and d are positive integers, then c 2 + d 2 must be at least 2. Thus c d = c 2 + d 2 1 is at least 1. (e) The set of rational numbers is closed under. Yes. If c and d are rational, then each of c 2 and d 2 must be rational. Since the sum and difference of two rationals is rational, it follows that c d = c 2 + d 2 1 is also rational. (f) The set of irrational numbers is closed under. No. Let c = d = 2. Then c and d are irrational (as proven in class). But c d = c 2 + d 2 1 = 5, which clearly is rational. 12. Let X = {0, 2, 5, 8, 13, 15} and Y = P(X). Define T: X P(X) as follows: Find D * (as defined in Part II, problem 1) { j} if j is prime j X T( j) {0, 8, 13} if j is not prime Answer: D * = {15} 13. Give an example of three sets, X, Y, S, and two mappings, F: X Y, G: Y S such that G is not injective yet G F is injective. Let X = {a}, Y = {b, c} and S = {d}.

7 Define F: X Y by: F(a) = c and G: Y S by G(b) = G(c) = d. Then G is clearly not injective, yet G F is injective since the domain of G F consists of a single point Let S be the set of all polynomials for which odd powers of x (i.e., 1, 3, 5, 7, ) do not appear. For example, 4 + ½ x 2 x 8 S, but x x81 S. (a) Is S closed under differentiation? Why? No. Counter-example. Let p(x) = x 4 ; this is a member of S but dp/dx = 3x 3 is not a member of S. (b) Is S closed under the operation of taking the second derivative? Yes. Let p be a finite sum of terms of the form k x n. where k is any real number and n is any even integer 2 Now d 2 /dx 2 p = kn(n-1) x n-2. But since n is even, n 2 must be even as well. (c) Is S closed under multiplication by x 3? No. Counter-example: Let p(x) = x 4. Then x 4 p(x) = x 7 S. (d) Is S closed under multiplication by x 4? Yes, x 4 p(x) consists of a finite sum of terms of the form x n+4. Since n is even, so is n+4. (e) Is S closed under the following operation *? For all p S p = p(1 + x 4 ) Yes. Let n be a non-negative even integer. Then one must verify that (1 + x 4 ) n consists only of even powers of x. This follows directly from the binomial theorem. (f) Is S closed under the following operation : For all p S p = p(1 + x) No. Counter-example. Let p(x) = x 2. Then p(1 + x) = (1 + x) 2 = 1 + 2x + x 2 S. Extra Credit: When 200! Is expanded, how many consecutive trailing zeroes appear at the end of the number?

8 200! may be written (uniquely) as the product abc where a is a power of 2, b is a power of 5 and c is neither divisible by 5 nor by 2. Now the number of occurrences of 10 is equal to the number of 5s in 2001! since the 2s are more abundant than the 5s. 8 Now we must count: There are 40 numbers in the product divisible by 5 8 numbers divisible by 25 1 number divisible by 125 Hence the number of 5s (and thus the number of terminating 0s) in 200! Is = 49.