Prove that if a statement can be proved by strong mathematical induction, then it can he proved by ordinary mathematical induction. To do this, let P( n ) be a property that is defined for each integer n , and suppose the following two statements are true: 1. P ( a ) , P ( a + 1 ) , P ( a + 2 ) , ... , P ( b ) . 2. For any integer k ≥ b , if P ( i ) is true for each integer i from a through k , then P ( k + 1 ) is true. The principle of strong mathematical induction would allow us to conclude immediately that P ( n ) is true for every integer n ≥ a . Can we reach the same conclusion using the principle of ordinary mathematical induction? Yes! To see this, let Q ( n ) be the property P ( j ) is true for each integer j with a ≤ j ≤ n . Then use ordinary mathematical induction to show that Q ( n ) is true for every integer n ≥ b . That is, prove: 1. Q ( b )is true. 2. For each integer k ≥ b , if Q ( k ) is true then Q ( k + 1 ) is true.
Prove that if a statement can be proved by strong mathematical induction, then it can he proved by ordinary mathematical induction. To do this, let P( n ) be a property that is defined for each integer n , and suppose the following two statements are true: 1. P ( a ) , P ( a + 1 ) , P ( a + 2 ) , ... , P ( b ) . 2. For any integer k ≥ b , if P ( i ) is true for each integer i from a through k , then P ( k + 1 ) is true. The principle of strong mathematical induction would allow us to conclude immediately that P ( n ) is true for every integer n ≥ a . Can we reach the same conclusion using the principle of ordinary mathematical induction? Yes! To see this, let Q ( n ) be the property P ( j ) is true for each integer j with a ≤ j ≤ n . Then use ordinary mathematical induction to show that Q ( n ) is true for every integer n ≥ b . That is, prove: 1. Q ( b )is true. 2. For each integer k ≥ b , if Q ( k ) is true then Q ( k + 1 ) is true.
Solution Summary: The author explains that if a statement can be proved by strong mathematical induction, then it can also be proven by ordinary mathematics.
Prove that if a statement can be proved by strong mathematical induction, then it can he proved by ordinary mathematical induction. To do this, let P(n) be a property that is defined for each integer n, and suppose the following two statements are true:
1.
P
(
a
)
,
P
(
a
+
1
)
,
P
(
a
+
2
)
,
...
,
P
(
b
)
.
2. For any integer
k
≥
b
, if P(i) is true for each integer i from a through k, then
P
(
k
+
1
)
is true. The principle of strong mathematical induction would allow us to conclude immediately that P(n) is true for every integer
n
≥
a
. Can we reach the same conclusion using the principle of ordinary mathematical induction? Yes! To see this, let Q(n) be the property P(j) is true for each integer j with
a
≤
j
≤
n
.
Then use ordinary mathematical induction to show that Q(n) is true for every integer
n
≥
b
. That is, prove:
1. Q(b)is true. 2. For each integer
k
≥
b
, if Q(k) is true then
Q
(
k
+
1
)
is true.
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