College Algebra 9th txtbk

516 CHAPTER 8 SEQUENCES, INDUCTION, AND PROBABILITY
P k
PART 2 Show that if is true, then P k1 is true.
P k : (xy) k x k y k
P k1
Assume is true.
P k1 : (xy) k1 x k1 y k1 P k
P k1
Show that follows from .
Here we start with the left side of and use to find the right side of :
(xy) k1 (xy) k (xy) 1
x k y k xy
(x k x)(y k y)
x k1 y k1
Use P k : (xy) k x k y k
Use properties of real numbers.
Use Definition 1 twice.
So (xy) k1 x k1 y k1 , and we have shown that if is true, then is true.
CONCLUSION
Both conditions in Theorem 1 are satisfied. Therefore,
P k
P k
P n
P k
P k1
P k1
is true for all positive integers n.
MATCHED PROBLEM 4 Prove that (x/y) n x n /y n for all positive integers n.
Example 5 deals with factors of integers. Before we start, recall that an integer p is
divisible by an integer q if p qr for some integer r.
EXAMPLE 5
Proving a Divisibility Property
Prove that 4 2n 1 is divisible by 5 for all positive integers n.
PROOF Use the definition of divisibility to state P n as follows:
PART 1 Show that
P 1
P n : 4 2n 1 5r for some integer r
is true.
P 1 : 4 2 1 15 5 3
So P 1 is true.
P k
PART 2 Show that if is true, then is true.
P k : 4 2k 1 5r
P k1 : 4 2(k1) 1 5s
P k1
As before, we start with the true statement P k :
4 2k 1 5r
4 2 (4 2k 1) 4 2 (5r)
4 2k2 16 80r
4 2(k1) 1 80r 15
5(16r 3)
So
4 2(k1) 1 5s
for some integer r Assume is true.
for some integer s Show that must follow.
Multiply both sides by 4 2 .
Simplify.
Add 15 to both sides.
Factor out 5.
where s 16r 3 is an integer, and we have shown that if is true, then is true.
CONCLUSION
Both conditions in Theorem 1 are satisfied. Therefore,
P k1
P n
P k
P k
P k1
P k1
is true for all positive integers n.