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Unformatted text preview: ”@0697 23\ Ht e M ms 3 '6 SECTION 1.7 Proof Methods and Strategy —— 2. The cubes that might go into the sum are 1, 8, 27, 64, 125, 216, 343, 512, and 729. We must show that
no two of these sum to a number on this list. Ifwe try the 45 combinations (1 + 1,1 +8, ..., 1 +729, 8 +8, 8 + 27, . . . 8 + 729, . . . , 729 + 729), we. see that none of them works. Having exhausted the possibilities, we
conclude that no cube less than 1000 is the sum of two cubes. _ 4. There are three main cases, depending on which of the three numbers is smallest. If a is smallest (or tied fOI
smallest), then clearly a S min(b, c) , and so the left—hand side equals a. On the other hand, for the right—hand
side we have min(a, c) = a as well. In the second case, b is smallest (or tied for smallest). The same reasoning
shows us that the righthand side equals b; and the lefthand side is min(a, b)_= b as well. In the ﬁnal case,
in which 0 is smallest (or tied for smallest), the left—hand side is min(a, c) = c, whereas the righthand side is
clearly also c. Since one of the three has to be smallest we have taken care of all the cases. 3H 28. There are two things to prove. For the “if” part, there are two cases. If m = n, then of course m = n ;
if m = —n, then m2 = (—77.)2 = (—1)2'n.2 = 77.2. For the “only if” part, we suppose that m2 = 7?. Putting
 . everything on the left and factoring, we have (m + n)(m — n) = 0. Now the only way that a product of two
numbers can be zero is if one'of them is zero. Therefore we conclude that either m + n = 0 (in which case m = —n), or else m — n = O (in which case m = n), and our proof is complete. L
i?) SECTION 2.1 Sets ,
V 2. There are of course an inﬁnite number of correct answers. ,1
a) {3nln=0,1,2,3,4} or {mlmisamultipleof$AOgmgl2}. . .
b) {m  —3 _<_ .7: g 3}, where we are assuming that the domain (universe of discourse) is the set of integers. c) {:z:  :1: is a letter of the word monopoly other than I or y}. ‘7 4. Each of the sets is a subset of itself. Aside from that, the only relations are B Q A, C Q A, and C g D. \, 6. a) Since the set contains only integers and {2} is a set, not an integer, {2} is not an element.
b) Since the set' contains only integers and {2} is a set, not an integer, {2} is not an element.
c) The set has two elements. One of them is patently {2}.
d) The set has two elements. One of them is patently {2}. e) The set has two elements. One of them is patently {2}. . . .
f) The set has only one element, {{2}}; since this is not the same as {2} (the former is a set containing a set, whereas the latter is a set containing a number), {2} is not an element of {{{2}}}. g 8. a) true b) true c) false—see part (a) d) true _
e) true—the one element in the set on the left is an element of the set on the right, and the sets are not equal f) true—similar to part (e) g) false—the two sets are equal @043! m Moe m "MB we can let A = G and B = {G} as the simplest example. ~18. The cardinality of a set is the number of elements it has.
a) The empty set has no elements, so its cardinality is 0.
b) This set has one element (the empty set), so its cardinality is 1.
c) This set has two elements, so its cardinality is 2.
d) This set has three elements, so its cardinality is 3. ‘ 16. Since the empty set is a subset of every set, we just need to take a set B that contains 0 as an element. Thus . 1f 26. We can conclude that A = Q or B): 6. To prove this, suppose that neither A nor B were empty. Then
there would be elements a E A and b E B. This would give at last one element, namely' (a, b), in A x B, so A x B would not be the empty set. This contradiction shows that either A or B (or both, it goes without saying) is empty. ' 30. Suppose A 9A B and neither A nor B is empty. We must prove that A x B 9A B x A. Since A $ B , either '
we can ﬁnd an element a: that is in A but not B, or vice versa. The two cases are similar,_so without loss of generality, let us assume that a: is in A but not B. Also, since B is not empty, there is some element y E B. Then (a), y) is in A x B by deﬁnition, but it is not in B x A since :1: 9E B. Therefore A x B gé B x A. ._ gm “  .nL1___~ “A J‘ru all nrnrﬁl‘al n11TDOSeS .In each case we want the set of all values of z or inequality, in the domain (the set of integers) gers, so the truth set is the em t
0) Negative integers certainly satisfy this inequality P y set, 0
’ andljélz. Thusthetruthsetis {mEZI$<$ i 3'1 SECTION 2.2 Set Operations
2. a)»AﬂB b)AﬂB,whichisthesameasA—B c)AUB d) AUB 4. Note that A g B.
a) {a,b,c,d,e,f,g,h}=B b) {a,b,c,d,e}=A
c) There are no elements in A that are not in B, so the answer is G. d) { f, g, h} as do all positive integers greater than 1. However, 0 7i 02 2}={xe,ziz¢QAz¢1}= {.. ,,—3 —,2—1,2,3,...}. Flochf 2?] HQ 6 l7? gGED 2.2.92. 1Way Containment Proof: Let XE AUB. Then XE A or XE B.
So at least one of the following two cases must apply: (i) Suppose X E A. Then since A g C, we know that X E C. '
But Exercise 16(b) implies that C ; CUD. So X E CUD. (ii) Suppose X E B. Then since B g D, we know that X E D.
1 But Exercise 16(b) implies that D g CUD. So X E CUD. So in any case, we have X E CUD.
Since X E AUB implies that X E CUD, we conclude that AUB ; CUD. I i4. Since A = (A — B) U (A 0 B), we conclude that A = {1,5,7,8} U {3,6,9} = {1,3,5,6,7,8,9}. Similarly
B = (B——A) U (AﬂB) 2: {2,10} U{3,6,9} = {2,3,6,9,10}. . a) If a: is in A m B, then perforce it is in A (by deﬁnition of intersection).
b) If a: is in A, then perforce it is in A U B (by deﬁnition of union).
c) If a: is in A — B, then perforce it is in A (by deﬁnition of difference).
d) If a: E. A then z ¢ B — A. Therefore there can be no elements in A 0 (B —— A), so A m (B —— A) = G.
e) The left—hand side consists precisely of those things that are either elements of A or else elements of B
but not A, in other words, things that are elements of either A or B (or, of course, both). This is precisely
the deﬁnition of the right—hand side. . a) Suppose that a: E A U B. Then either a: E A or a: E B. In either case, certainly a: E A UB U C'. This establishes the desired inclusion. b) Suppose that :1: E A 0 B 0 0. Then :1: is in all three of these sets. In particular, it is in both A and B and therefore in A m B, as desired. c) Supposethat :1: E (A—B) —C'. Then :1: isin A—B but not in 0. Since :1: E A~B, we knowthat :1: E A (we also know that :1: ¢ B, but that won’t be used here). Since we have established that :1: E A but :1: ¢ 0, we have proved that :1: E A — C' . ‘ ' d) To show that the set given on the left—hand side is empty, it sufﬁces to assume that :1: is some element in that set and derive a contradiction, thereby shovV'mg that no such :1: eXists. So suppose that :1: E (A — C') m (C' — B) .
i Then :1: E A — C' and :1: E C' — B. The ﬁrst of these statements implies by deﬁnition that :1: ¢ 0, while the
second implies that :1: E C' . This is impossible, so our proof by contradiction is complete.
E e) To establish the equality, we need to prove inclusion in both directions. To prove that (B — A) U (C' — A) Q
(B U 0) ~ A, suppose that :1: E (B — A) U (C' — A). Then either :1: E (B — A) or :1: E (C' —— A). Without loss of
generality, assume the former (the proof in the latter case is exactly parallel.) Then :1: E B and :1: ¢ A. From
’ the ﬁrst of these assertions, it follows that :1: E B U 0. Thus we can conclude that a: E (B U C') — A, as desired.
For the converse, that is, to show that (B U C') — A Q (B — A) U (C' — A), suppose that z E (B U C') — A.
This means that z E (B U C') and :1: ¢ A. The ﬁrst of these assertions tells us that either :1: E B or :1: E C' .
Thus either :1: E B — A or :1: E C' — A. In either case, :1: E (B — A) U (C’ — A). (An alternative proof could be
given by using Venn diagrams, showing that both sides represent the same region.) I
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