1.5(8)
a)If we use modus tollens starting from the back, then we conclude that I am not sore. Another application of modus tollens then tells us that I did not play hockey.
b)We really can’t conclude anything specific here.
c)By universal instantiation, we conclude from the first implication by modus ponens that dragonflies have six legs, and we conclude by modus tollens that spiders are not insects. We could say using existential generalization that, for example, there exists a non-six-legged creature that eats a six-legged creature, and that there exists a non-insect that eats an insects.
d)We can apply universal instantiation to the implication and conclude that if Homer(respectively, Maggie) is a student, then he(she) has an Internet account. Now modus tollens tells us that Homer is not a student. There are no conclusions to be drawn about Maggie.
e)The first implication is that if x is healthy to eat, then x does not taste good. Universal instantiation and modus ponens therefore tell is that tofu does not taste good. The third sentence says that if you eat x, then x tastes good. Therefore the fourth hypothesis already follows (by modus tollens) from the first three. No conclusions can be drawn about cheeseburgers from these statements.
f)By disjunctive syllogism, the first two hypotheses allow us to conclude that I am hallucinating. Therefore by modus ponens we know that I see elephants running down the road.
1.5(14)
a)This is not valid, because the converse of an implication dies not follow from the implication(it’s the fallacy of affirming the conclusion).
b)This argument is valid, using universal instantiation and modus ponens.
1.5(22)
a)We must prove the contrapositive: If n is odd, then 3n + 2 is odd. Assume that n is odd. Then we can write n = 2k + 1 for some integer k. Then 3n + 2 = 3 ( 2k + 1 ) + 2 = 6k + 5 = 2 ( 3k + 2 ) + 1. Thus 3n + 2 is two times some integer plus 1, so it is odd.
b)Suppose that 3n + 2 is even and that n is odd. Since 3n + 2 is even, so is 3n. If we add subtract an odd number from an even number, we get an odd number, so 3n – n = 2n is odd. But this is obviously not true. Therefore our supposition was wrong, and the proof by contradiction is complete.
1.5(34)
There are three main cases, depending on which of the three numbers is smallest. If a is smallest (or tied for smallest), then clearly a <= min(b,c), and so the left-hand side equals a. On the other hand, for the right-hand side we gave min(a,c) = a as well. In the second case, b is smallest (or tied for smallest). The same reasoning shows us that the right-hand side equals b; and the left-hand side is min(a,b) = b as well. In the final case, in which c is smallest (or tied for smallest), the left-hand side is min(a,c) = c, whereas the right-hand side is clearly also c. Since one of the three has to be smallest we have taken care of all the cases.
1.5(54)
We know from algebra that the following equations are equivalent: ax + b = c, ax = c - b. x = ( c – b ) / a . This shows, constructively, what the unique solution of the given equation is.
1.6(4)
Each of the sets is a subset of itself. A side from that, the only relations are and.
1.6(14)
a)0
b)1
c)2
d)3
1.6(20)
By the definition it is the set of all ordered pairs such that is a course and is a professor.
1.7(12)
a)If x is in, then perforce it is in A. (by definition of intersection)
b)If x is in A, then perforce it is in. (by definition of union)
c)If x is in, then perforce it is in A. (by definition of difference)
d)Ifthen. Therefore there can be no elements in, so.
e)The left-hand side consist precisely of those things that are either elements of Aor 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.
1.7(18)
First suppose x is in the left-hand side. Then x must be in A but in neitherB nor C. Thus , but , so x is in the right-hand side. Next suppose that x is in the right-hand side. Thus x must be in and not in. The first of these implies that and. But now it must also be the case that, since otherwise we would have. Thus we have shown that x is in A but in neitherB nor C, which implies that x is in the left-hand side.
1.7(36)
To count the elements of we proceed as follows. First we count the elements in each of the sets and add. This certainly gives us the elements in the union, but we have overcounted. Each element in,, and has been counted twice. Therefore we subtract the cardinalities of these intersections to make up for the overcount. Finally, we have compensated a bit too much, since the elements of have now been counted three times and subtracted three times. We adjust by adding back the cardinality of.
1.8(8)
a)1
b)2
c)-1
d)0
e)3
f)-2
g)1
h)2
1.8(12)
a)Yes
b)No
c)Yes
d)No
1.8(16)
a)
b)
c)
d)
1.8(22)
a)
b)
c)
d)