epdf.pub_the-nuts-and-bolts-of-proofs-third-edition-an-intr

Special Kinds of Theorems 37
Parti. If A, then B.
The fact that the number n is odd is sufficient to imply that its square is odd.
By hypothesis the number n is odd. So we can write n = 2p-\-l, where p is an
integer number. Therefore,
n^ = (2p-\-lf
= 4/?^ + 4p + 1
= 2(2p2 + 2p)4-l.
Because p is an integer number, the number s = 2p^ -\- Ip is integer as well.
Thus,
This proves that r? is odd.
n^ = 2s+\.
Part 2. If B, then A.
The fact that the square of number n is odd is sufficient to imply that the
number itself is odd.
Discussion: If we know that n^ is odd, we can only write n^ = 2t-j-l, with t
positive integer. We cannot write that n^ = (2k + 1)^, because this is the
conclusion we are trying to reach. If n^ = 2t + 1, then
n = y/n^ = V2r+1.
This equahty does not give us any useful information. So, we need to look
for another starting point. We can try to prove its contrapositive. Let us
assume "not A"; that is, the number n is not odd.
Because n is an even number, it can be written as n = 2t, with t positive
integer; therefore, n^ = 4t^. This imphes that n^ is even, as we can write it as
n^ = 2(2t^X and 2t^ is an integer number.
Thus, we have proved that "not A" imphes "not B." So, the statement
"If B, then A" is true. •
As can be seen from Example 2, the proof of an equivalence theorem might
require the use of different techniques (e.g., direct proof and use of the
contrapositive) for the different parts of the proof
Some theorems list more than two statements and claim that they are all
equivalent. The construction of the proof of these theorems is rather flexible
(that is, it can be set up in several ways), as long as we establish that each
statement implies each of the other statements and that each statement is
implied by each of the other statements. In this way we prove that each
statement is sufficient and necessary for all the others. Some of the