Q8 of 27 Page 4

Prove that 2n > n for all integers n.

Let P(n) = 2n > n


For n =1,


21 > 1.


Hence P (1) is true.


Assume that P(k) is true for any positive integers k, i.e.


2k > k …… (1)


We will now prove that P(k+1) is true whenever P(K) is true.


Multiplying both sides of (1) by 2, we get


2. 2k > 2 k


⇒ 2k+1 > 2k = k + k > k + 1


Hence P(k+1) is true whenever P(K) is true.


By mathematical induction P(n) is true for all n.


More from this chapter

All 27 →
6

If P(n) is the statement “2n ≥ 3n”, and if P(r) is true, prove that P(r + 1) is true.

 7

If P(n) is the statement “n3 + n is divisible by 3”, prove that P(3) is true but P(4) is not true.

 9

For every positive integer n, prove that 7n – 3n is divisible by 4.

 10

Using the principle of mathematical induction, prove that (23n – 1) is divisible by 7 for all n ∈ N.