Math 105A HW 1 Solutions

Size: px

Start display at page:

Download "Math 105A HW 1 Solutions"

Peter West
5 years ago
Views:

1 Sect : # 2, 3 (Page 7-8 Math 105A HW 1 Solutions 2(a ( Statement: Each positive integers has a unique prime factorization. n N: n = 1 or ( R N, p 1,..., p R P such that n = p 1 p R and ( n, R, S N, q 1,..., q R, p 1,..., p S P: If n = q 1 q R = p 1 p S, then (R = S and permutation σ S R such that p 1 = q σ(1,..., p R = q σ(r. Negation: There exists a positive integer which has no prime factorization or ( for which the prime factorization is not unique. n N : n 1 and ( R N, p 1,..., p R P : n p 1 p R or ( n, R, S N, q 1,..., q R, p 1,..., p S P: n = q 1 q R = p 1 p S and (R S or permutations σ S R : i {1,..., R} : p i = q σ(i. Note: Here we use P to denote the set of all prime numbers, and we use S R to denote the set of all permutations of {1,..., R}. Note that a permutation is a bijective function of {1,..., R} onto itself. 2(b Statement: ( 2 is the only even prime number. 2 P and n P : If n is even, then n = 2. Negation: 2 is not a prime number or there exists another even prime number besides ( 2. 2 / P or n P: n is even and n 2. 2(c Statement: Multiplication (in N is associative. a, b, c N : a(bc = (abc. Negation: Multiplication (in N is not associative. a, b, c N : a(bc (abc. 2(d Statement: Two points in the plane determine a line. points P 1, P 2 line L : P 1 L and P 2 L Negation: There exists two points in the plane which do not both lie on a line. points P 1, P 2 lines L : P 1 / L or P 2 / L 2(e Statement: The altitudes of a triangle intersect in a point. triangles ABC point P : P lies on each of the three altitudes of ABC Negation: There is a triangle in which the altitidudes do not intersect in a common point. triangle ABC points P : P does not lie on all three altitudes of ABC

2 2(f Statement: Given a line in a plane and point not on it, there exists a unique line passing through the given point parrallel to the given line. lines L, points P : P / L {( line L 1 : L 1 L and P L 1 and } ( lines L 1, L 2 : L 1 L and L 2 L and P L 1 and P L 2 L 1 = L 2 Negation: Given a line in a plane and point not on it, there does not exist a line passing through the point and being parallel to the line, or there exist two distinct lines with this property. line L, point P : P / L and {( line L 1 : L 1 L or P / L 1 } or ( lines L 1, L 2 : L 1 L and L 2 L and P L 1 and P L 2 and L 1 L 2 2(f Statement: Any partitioning of the integers into a finite number of disjoint subsets has the property that one of the subsets contains arbitrarily long arithmetic progressions R 1, S 1,..., S R Z : Sr = N, S r S s = r {1,..., R} : l N n 0 Z, p N: {n 0, n 0 + p, n 0 + 2p,..., n 0 + lp} S r Negation: There exists a partitioning of the integers into a finite number of disjoint subsets such that none of the subsets contains arithmetic progressions of arbitrary length. R 1, S 1,..., S R Z : Sr = N, S r S s = and r {1,..., R} : l N n 0 Z, p N: {n 0, n 0 + p, n 0 + 2p,..., n 0 + lp} S r 2(f Statement: If there are more letters than mailboxes, at least one mailbox must have two letters. (disjoint sets M 1,..., M R : M 1... M R > R r {1,..., R} : M r 2 Negation: There is a distribution of letters into mailboxes, such that each mailbox contains at most one letter, yet the number of letters is bigger than the number of mailboxes. (disjoint sets M 1,..., M R : M 1... M R > R and r {1,..., R} : M r 1 #3(a Every line segment has a midpoint: line segments AB point P : P is the midpoint of AB point P AB : P is the midpoint of AB (There exists a point which is the midpoint of every line segment False.

3 #3(b Every nonzero rational number has a reciprocal. a Q \ {0} b Q : ab = 1 b Q a Q \ {0} : ab = 1 (There exists a number b which is the reciprocal of every nonzero number False. #3(c Every non-empty set of positive integers has a smallest element. A N, A a N: a is the smallest element of A. a N A N, A : a is the smallest element of A. (There exists an element which is the smallest element of every nonempty subset A of N false. #3(d There is no largest prime. p P q P : q > p q P p P : q > p (There exists a prime such that any other prime is smaller false. Remark: there might be other valid solutions depending on how the original statemente is expresses in terms of quatifiers. Sect : # 1, 2, 3, 4, 5, 6, 7* (Page Prove that every subset of N is either finite or countable. Suppose that A is a subset of N. What we need to do is to show that we have a way to count all the elements in A. One way to count all elements in A is to count them in their order from small to large. But, how do we find the smallest each time? We recall and use the fact that each non-empty subset of N has a smallest element. We let a 1 be the smallest element of A. Then remove it and let a 2 be the smallest element of the remaining set A \ {a 1 }. Then remove also a 2 and let a 3 be the smallest element of A \ {a 1, a 2 }. And so on. Notice that we can make the (n + 1-th step as long as A \ {a 1,..., a n } is not empty. We thus end up with two cases: either the process stops at some point or we can continue indefinitely. Consequently we will either get finitely many elements {a 1,..., a n } or an infinite sequence of them {a 1, a 2,...}. Clearly, a 1 < a 2 <... and also a k k. In the first case, we have = A \ {a 1,..., a n } for some n and thus A = {a 1,..., a n } is finite. In the second case we extracted from A infinitely many elements. But did we really get all of them? Yes, but this is somewhat tricky: Suppose there is an m A which is different from all the a n s. As a m m we find a minimal n such that a n m. This means either a 1 m or a n m > a n 1. Now

4 consider the n-th step where we defined a n as the minimum of A\{a 1,..., a n 1 }. But since m a n < a n+1 <... we either have m = a n or a n is not the minimum of the remaining set (which is a contradiction. Thus m is equal to one of the a n s and we can conclude that A = {a 1, a 2,...} = {a n : n N}. i.e., A is countably infinite. That is we have a one-to-one correspondence between the set of natural numbers and A. Therefore any infinite subset A of the set of natural numbers is countable, i.e has the cardinality N. 2. The set F of all finite subsets of N is countable. First think what are the finite subsets of natural numbers: A = {a 1, a 2,, a k } The number of elements in A is k, which is finite. Therefore all possible finite subsets are precisely the A s as above with k = 0, 1, 2,. In other words, the set of all finite subset is F = k=0 Card(A=k A N {A}. To apply the Theorem that states: a countable union of countable sets is countable, we only need to show that for each k the set is countable. But {A N : Card(A = k} {A N : Card(A = k} = N N N }{{} k times is the Cartesian product of k copies of N, which is countable by the Theorem that states: Cartesian product of countable sets is countable. There is also another proof. We can observe that F = P({1,..., n}, k=1 which just says that for each finite subset A N there exists an n N such that A {1,..., n}. Indeed, we may take for n the largest element in A. The power set P({1,..., n} ha precisely 2 n elements. Thus we have written F as the countable union of finite sets, which is countable by a known theorem. On the other hand, it is clear that F is infinite.

5 3. The rational numbers are countable. Write Q = k N Q k as a countable union of sets Q k = { } j k : j Z. Each of these sets itself is countable since there is a bijective function f k : Z Q k, f k (j = j/k, and Z has been shown to be countable earlier. Now we apply the Theorem which states: a countable union of countable sets is countable. (Notice that the Q k are not disjoint, but the theorem applies nonetheless. Thus Q is countable, while it is clear that Q is infinite. Another proof would consist in arranging all the rational numbers in an array (take care of the negative ones!, crossing out all duplicates, and using Cantor s first diagonal argument in order to denumerate them. 4. If a countable set is removed from an uncountable set, the remainder is uncountable. Suppose that A is a uncountable set and C = A \ B where B is a countable set. We prove by contradiction. I.e., assume that C is countable. Then we conclude from A = C B where both B and C are countable, that A is countable, which contradicts with the assumption that A is uncountable. 5. A Cartesian product of countably many sets A 1, A 2,..., each of which contains at least two elements, is uncountable. One possible proof: Let B k be subsets of A k containing precisely two elements. Then B 1 B 2 B 3... A 1 A 2 A Hence it is sufficient to show that the Cartesian product of the B k s is uncountable. (Why? Argue by contradiction and use the fact that any subset of a countable set is countable or finite. However, now it is easy to make a bijection between the Cartesian product of the B k s and the set of all sequences {x n } n=1 with x n {0, 1}, which we know it is uncountable. A second proof: Just modify Cantor s second diagonal argument. 6. Let A be a set for which there exists a function f : A N such that for every k the preimage f 1 (k is finite. Show that A is countable or finite. For k N define the set A k = {a A : f(a = k}. By assumption, each A k is finite subset of A. On the other hand, we have A = A k. k=1 It is obvious that holds. For the converse, let a A. Then put k = f(a (which one can do since A is the domain of definition of A. The definition of A k implies that a A k, i.e., a is in the union of all A k s. To complete the argument

6 we notice that the countable union of countable (or finite sets is countable or finite. Remark: It can indeed happen that A is finite. This happens precisely if all but finitely many of the A k s are empty. 7. The power set 2 A has a bigger cardinality as A. First of all, we can identify a A {a} 2 A which implies that card(a card(2 A. (We have an injective function f : A 2 A. To show that card(a < card(2 A we only need to show that card(a card(2 A. Again, we prove by contradiction. We assume that card(a = card(2 A. That means there exists a one-to-one correspondence between A and 2 A. In other words, we can label each element in 2 A by elements in A, that is, 2 A = {S a : a A}. (The map a A S a 2 A is the corresponding bijective function. Now, we will get a contradiction by constructing a subset of A using the above labeling, which has to be without label. Let S = {a A : a / S a }. Clearly S 2 A, that is to say, S = S a0 for some a 0 A. But this creates a contradiction: if a 0 S a0, then, by the definition of S = S a0, a 0 / S a0 ; if a 0 / S a0, then, by the definition of S = S a0, a 0 S a0. Either way things do not add up. This means S S a0 for any a 0 A. But that contradicts with our assumption that every element in 2 A has a label by A. Thus card(a card(2 A.

1.4 Cardinality. Tom Lewis. Fall Term Tom Lewis () 1.4 Cardinality Fall Term / 9

1.4 Cardinality. Tom Lewis. Fall Term Tom Lewis () 1.4 Cardinality Fall Term / 9 1.4 Cardinality Tom Lewis Fall Term 2006 Tom Lewis () 1.4 Cardinality Fall Term 2006 1 / 9 Outline 1 Functions 2 Cardinality 3 Cantor s theorem Tom Lewis () 1.4 Cardinality Fall Term 2006 2 / 9 Functions