Answer
Have to prove
a)P(1, 1) is true and P(n, k) → [P(n + 1, k) ∧ P(n, k + 1)] is true for all positive integers n and k.
b) P(1, k) is true for all positive integers k, and P(n, k) → P(n + 1, k) is true for all positive integers
n and k.
c) P(n, 1) is true for all positive integers n, and P(n, k) → P(n, k + 1) is true for all positive integers n and k.
Work Step by Step
--In each case,
- give a proof by contradiction based on a “smallest counterexample,” that is, values of n and k such that P(n, k) is not true and n and k are smallest in some sense.
a)
- -Choose a counterexample with n + k as small as possible.
We cannot have n = 1 and k = 1,
-because we are given that P(1, 1) is true.
Therefore, either n > 1 or k > 1. In the former case, by our choice of counterexample,
we know that P(n−1, k) is true. But the inductive step then forces P(n, k)
to be true, a contradiction.The latter case is similar.
So our supposition that there is a counterexample mest be wrong,
and P(n, k) is true in all cases.
b)
-- Choose a counterexample
with n as small as possible.
- We cannot have n = 1,
- because we are given that P(1, k) is true for all k. Therefore, n > 1.
By our choice of counterexample,
- we know that P(n − 1, k) is true.
But the inductive step then forces P(n, k) to be true, a contradiction.
c)
--Choose a counterexample with k as small as possible.
We cannot have k = 1, because we are given that P(n, 1) is true for all n. Therefore, k > 1.
- By our choice of counterexample,
- we know that P(n, k − 1) is true.
But the inductive step then forces P(n, k) to be true, a contradiction.
