By the principle of mathematical induction, prove 2n > n, for all n �

1.	By the principle of mathematical induction, prove 2n > n, for all n ∈ N.
Answer» Let P(n) denote the statement 2ⁿ > n for all n ∈ N i.e., P(n): 2ⁿ > n for n ≥ 1 Put n = 1, P(1): 2¹ > 1 which is true. Assume that P(k) is true for n = k i.e., 2^k > k for k ≥ 1 To prove P(k + 1) is true. i.e., to prove 2^{k + 1} > k + 1 for k ≥ 1 Since 2^k > k Multiply both sides by 2 2 . 2^k > 2k 2^{k + 1} > k + k i.e., 2^{k + 1} > k + 1 (∵ k ≥ 1) ∴ P(k + 1) is true whenever P(k) is true. ∴ By principal of mathematical induction P(n) is true for all n ∈ N.

By the principle of mathematical induction, prove 2n > n, for all n ∈ N.

Answer»

Let P(n) denote the statement 2ⁿ > n for all n ∈ N

i.e., P(n): 2ⁿ > n for n ≥ 1

Put n = 1, P(1): 2¹ > 1 which is true.

Assume that P(k) is true for n = k

i.e., 2^k > k for k ≥ 1

To prove P(k + 1) is true.

i.e., to prove 2^{k + 1} > k + 1 for k ≥ 1

Since 2^k > k

Multiply both sides by 2

2 . 2^k > 2k

2^{k + 1} > k + k

i.e., 2^{k + 1} > k + 1 (∵ k ≥ 1)

∴ P(k + 1) is true whenever P(k) is true.

∴ By principal of mathematical induction P(n) is true for all n ∈ N.