Selected Solutions to Even Problems, Part 3

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Rosamund Morgan
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1 Selected Solutions to Even Problems, Part 3 March 14, 005 Page If one selects 101 integers from among {1,,..., 00}, then at least two of these numbers must be consecutive, and therefore coprime (which means that gcd(m, n) = 1). To see that there must be such a consecutive pair, we write the integers {1,,..., 00} as a union of 100 disjoint sets {1, } {3, 4} {5, 6} {199, 00}. Then, if we select 101 numbers, since there are only 100 such sets, we must have selected two numbers from the same set; that is, we must have selected a pair of consecutive numbers. 8. a. If S is a subset of the positive integers having at least 3 elements, we may write S = S e S o, where S e is the set of even elements belonging to S, and S o is the set of odd elements belonging to S. One or the other of these sets S e or S o must have at least two elements; else, S has at most elements. If S e has at least two elements, say x and y, then x + y is even; and, if S o has at least two elements, say x and y, then x + y is even. So, S contains at least two elements x and y such that x + y is even when S 3. b. If S has at least 5 elements, then there exists such a pair (x 1, y 1 ), (x, y ) S with (x 1, y 1 ) + (x, y ) = (z 1, z ), where z 1 and z are both even. To see that S must have at least 5 elements, consider the set S = {(1, 1), (1, ), (, ), (, 1)}. No sum of a pair of distinct elements of S will produce an ordered pair with both coordinates even. In general, any such set S may be partitioned as S = S e,e S e,o S o,e S o,o, 1

2 where S e,e is the set of ordered pairs (x 1, x ) S where both x 1 and x are even. The other sets are defined analogously. Now, if S has at least 5 elements, by the pigeonhole principle, one of these four disjoint sets will contain at least two elements. For example, S e,o may contain at least two elements (x 1, x ), (x 3, x 4 ). If so, then (x 1, x ) + (x 3, x 4 ) will have both coordinates even. The same is true for the sum of two elements from the other three sets S e,e, S o,e, ands o,o. So, if S has at least 5 elements, there is a pair of distinct elements of S who sum has even coordinates. c. Extending the idea from part b, if S Z + Z + Z +, and if S 9, then S has a pair who sum has even coordinates. Page 88.. a. On the interval (, 7], g(x) = x 8 x + = (x )(x + ) x + = (x ) = f(x). b. If we change A to ( 7, ], then f(x) g(x); in fact, f(x) is not even defined at x =. 10. a. f is invertible with inverse f 1 = {(x, y) : y + 3x = 7}. b. f is not invertible in general. For example, if a = c = 0 and b = 1, then f(x) = 0, which is not injective, so cannot be invertible (invertible bijection). c. f is invertible with inverse f 1 = {(x, y) : y = x 1/3 }. d. f is not invertible, since f(0) = f( 1) = 0, which implies f is not injective. (Recall f invertible f is a bijection) Page If R is an antisymmetric relation on a set with n elements, then for any pair of distinct elements a, b A we cannot have that both (a, b) and (b, a) belong to R, else a = b (so, they were not distinct elements after all.) Thus, the largest that R can be is if it contains all the pairs (a, a), together with one or the other of (a, b) or (b, a) for all pairs of distinct elements a, b A; so, ( ) n R n +.

3 An example of an antisymmetric relation where this bound on R is attained is A = {1,,..., n} and where the relation R is less-than-or-equal; that is, (a, b) R if and only if a b. Page If R is reflexive, then (a, a) R for all a A. Now, sice for every a A we have that there exists b A such that (a, b) A and (b, a) A, namely b = a, we have that (a, a) R, by the definition of the composition of two relations. It follows that R is also reflexive. 1. a and b. Suppose that R is a relation on a set A having n elements, which we can suppose are A = {1,,..., n}. Then, M(R) will be the matrix whose i, j entry is 1 if and only if (i, j) R; otherwise, the i, j entry is 0. So, this matrix is the 0 matrix if and only if R is empty. Part b is now obvious. c. From problem 11 we deduce that M(R n )M(R) = M(R n+1 ). Thus, for n =, M(R) = M(R ). Assume, for proof by induction, that M(R) k = M(R k ). Then, M(R k+1 ) = M(R k R) = M(R k )M(R) = M(R) k M(R) = M(R) k+1. The induction step is proved, and so the claim follows by mathematical induction. Here, note that the product of two matrices is defined in a funny way: M 1 M is the matrix whose i, j entry is 1 if and only if the i, j entry in the usual product of the matrices M 1 and M is 1; and the i, j entry in M 1 M is 0 otherwise. Page Suppose the poset (A, R) has two least elements, a 1 and a. Thus, a 1 b and a b for all b A. In particular, a 1 a and a a 1. Now, since partial orders are antisymmetric, it follows that a 1 = a, and therefore the least element, if it exists, is unique. 1. If R is a total relation if and only if the graph G has n + ( n ) edges. The term n comes from the edges corresponding to (a, a) R; and, the term ( ) n corresponds to the fact that the graph G contains edges corresponding to all 3

4 pairs of distinct vertices. Note that this directed graph G is not the Hasse diagram corresponding the the poset. Page a. The only element not accouted for in the conditions 1, A 1, 3, 4 A, 5, 6, 7 A 3 is the element 8, which can be in any of the three sets A 1, A or A 3. So, there are 3 possibilities for A 1, A, and A 3. b. If A 1 = 3, then A 1 contains either 7 or 8, but not both. One of the remaining sets A or A 3 must contain the remaining element. So, there are 4 possibilities (7 A 1, 8 A ; or 7 A 1, 8 A 3 ; or 8 A 1, 7 A ; or 8 A 1, 7 A 3 ). c. Either 7, 8 are both in the same set, which accounts for three possibilities for A 1, A, A 3 ; or, 7 is in one set, and 8 is in the other, which accouts for 6 more possibilies for A 1, A, A 3. So, there are 9 possibilities in all. Page We let c 1, c, c 3, c 4 be the conditions that a person brings hot dogs, chicken, salads, and dessert, respectively. So, we are given N = 65, N(c 1 ) = 1, N(c ) = 35, N(c 3 ) = 8, N(c 4 ) = 3, N(c 1 c ) = 13, N(c 1 c 3 ) = 10, N(c 1 c 4 ) = 9, N(c c 3 ) = 1, N(c c 4 ) = 17, N(c 3 c 4 ) = 14, N(c 1 c c 3 ) = 4, N(c 1 c c 4 ) = 6, N(c 1 c 3 c 4 ) = 5, N(c c 3 c 4 ) = 7, and N(c 1 c c 3 c 4 ) =. a. The people who set up and clean is N(c 1 c c 3 c 4 ), which is = 4. b. Those that bring only hotdogs are N(c 1 c c 3 c 4 ), which is N(c 1 ) N(c 1 c ) N(c 1 c 3 ) N(c 1 c 4 )+N(c 1 c c 3 )+N(c 1 c c 4 )+N(c 1 c 3 c 4 ) N(c 1 c c 3 c 4 ), which is c. Let =. S m = {c i1,...,c im } {c 1,...,c 4 } N(c i1 c im ); that is, for example if m =, then S m is N(c 1 c ) + N(c 1 c 3 ) + + N(c 3 c 4 ). We have that S 1 = 116, S = 75, S 3 =, S 4 =. 4

5 Then, the number of people who serve exactly one item is S 1 S + 3S 3 4S 4 = = 4. 5

MATH 215 Sets (S) Definition 1 A set is a collection of objects. The objects in a set X are called elements of X.

MATH 215 Sets (S) Definition 1 A set is a collection of objects. The objects in a set X are called elements of X. Notation 2 A set can be described using set-builder notation. That is, a set can be described