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

Special Kinds of Theorems 83
Let us construct the proof for Example 1 using the contrapositive of the
original statement, just to start becoming more famiHar with this kind of
statement:
"If the number b is not a multiple of 10, then either b
is not a multiple of 2 or fc is not a multiple of 5".
Proof: The two parts of the conclusion are "b is not a multiple of 2"
and "b is not a multiple of 5." To prove that the conclusion is true, it is
enough to prove that at least one of the two parts is true. (Keep reading
for more details regarding this kind of statement.)
Assume that the number b is not a multiple of 10. Then, by the division
algorithm,
b=10q-\-r
where q and r are integers and 1 < r < 9.
If r is an even number (i.e., 2, 4, 6, 8), then we can write r = It, with t
positive integer, and 1 < t < 4. So,
b=\Qq + 2t = 2(5q + 0-
The number 5^ +1 is an integer, so the number b is divisible by 2. But b is
not divisible by 5 because r is not divisible by 5.
Thus, in this case the conclusion is true because its second part is true.
If r is an odd number (i.e., 3, 5, 7, 9), then b is not divisible by 2. In this
case, the conclusion is true as well because its first part is true. •
We will now consider statements of the form "If A or B, then C."
In this kind of statement, we know that the hypothesis "A or B" is true.
This can possibly mean that:
1. Part A of the statement is true,
2. Part B of the statement is true,
3. Both parts A and B are true.
Because we do not know which one of the three cases to consider, we must
examine all of them. It is important to notice that it is sufficient to
concentrate on the first two cases, because the third case is a stronger case
that combines the first two. Therefore, the proof of a statement of the form
"If A or B, then C" has two parts (two cases):
1. Case 1. "If A, then C."
2. Casel "IfB, thenC."