Discrete Mathematics- An Open Introduction - 3rd Edition, 2016a

Selected Solutions 371
(b) The converse in words is this: for any number x, if everything times
x is zero, then everything added to x gives itself. Or in symbols:
∀x(∀z(x · z 0) → ∀y(x + y y)). The converse is true: the only
number which when multiplied by any other number gives 0 is x 0.
And if x 0, then x + y y.
(c) The contrapositive in words is: for any number x, if there is some
number which when multiplied by x does not give zero, then there
is some number which when added to x does not give that number.
In symbols: ∀x(∃z(x · z 0) → ∃y(x + y y)). We know the
contrapositive must be true because the original implication is true.
(d) The negation: there is a number x such that any number added to x
gives the number back again, but there is a number you can multiply
x by and not get 0. In symbols: ∃x(∀y(x + y y) ∧ ∃z(x · z 0)). Of
course since the original implication is true, the negation is false.
3.3.8.
(a) (¬P ∨ Q) ∧ (¬R ∨ (P ∧ ¬R)).
(b) ∀x∀y∀z(z x + y ∧ ∀w(x − y w)).
3.3.9.
(a) Direct proof.
Proof. Let n be an integer. Assume n is odd. So n 2k + 1 for some
integer k. Then
7n 7(2k + 1) 14k + 7 2(7k + 3) + 1.
Since 7k + 3 is an integer, we see that 7n is odd.
qed
(b) The converse is: for all integers n, if 7n is odd, then n is odd. We will
prove this by contrapositive.
3.3.10.
Proof. Let n be an integer. Assume n is not odd. Then n 2k for
some integer k. So 7n 14k 2(7k) which is to say 7n is even.
Therefore 7n is not odd.
qed
(a) Suppose you only had 5 coins of each denomination. This means
you have 5 pennies, 5 nickels, 5 dimes and 5 quarters. This is a total
of 20 coins. But you have more than 20 coins, so you must have more
than 5 of at least one type.