Section 6.2 A More General Principle of Mathematical Induction
What did I find difficult?
The Principle of Mathematical Induction
For a fixed integer m, let S = {i in Z: i > m} For each integer n in S, let P(n) be a statement. If
1) P(m) is true and
2) the implication If P(k) then P(k+1) is true for every integer k in S,
then P(n) is true for every integer n in S.
I'm beginning to understand induction, but I'm still having a tough time with it - mainly with the applications. It makes sense to me to prove that the smallest one is true, and then to prove the "domino effect" is true. I understand the concept - if one is true, then the next one is true - but it's hard for me to prove that in different cases.
What did I find interesting?
I really like the thought of induction - I can see how it is very useful. In looking through the results and their proofs, I can see the genius in applying the induction thought process. I'm excited to continue learning about it.
I really like the simple proof in result 6.9.
What did I find difficult?
The Principle of Mathematical Induction
For a fixed integer m, let S = {i in Z: i > m} For each integer n in S, let P(n) be a statement. If
1) P(m) is true and
2) the implication If P(k) then P(k+1) is true for every integer k in S,
then P(n) is true for every integer n in S.
I'm beginning to understand induction, but I'm still having a tough time with it - mainly with the applications. It makes sense to me to prove that the smallest one is true, and then to prove the "domino effect" is true. I understand the concept - if one is true, then the next one is true - but it's hard for me to prove that in different cases.
What did I find interesting?
I really like the thought of induction - I can see how it is very useful. In looking through the results and their proofs, I can see the genius in applying the induction thought process. I'm excited to continue learning about it.
I really like the simple proof in result 6.9.
"For every nonnegative integer n, 2^n > n. We prove that the inequality holds for n=0 since 2^0 > 0. Assume that 2^k > k, where k is a nonnegative integer. We show that 2^(k+1) > k+1. When k=0, we have 2^(k+1) = 2 >1 = k + 1. We therefore assume that k > k + 1.
By the Principle of Mathematical Induction, 2^n > n for every nonnegative integer n."
I can see how the simple inductions work. I am nervous to see how I do with the more difficult inductions. But I look forward to it.
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