Let Pn denote the statement that 7n 0 Base case Show P0 hol

Let P(n) denote the statement that 7·n = 0.

Base case: Show P(0) holds.

Since 7 · 0 = 0, P(0) holds.

Inductive step: Assume 7·j = 0 for all natural numbers j where 0 j k (induction hypothesis). Show P(k + 1): 7·(k + 1) = 0 .

Write k + 1 = i + j, where i and j are natural numbers less than k + 1.
Then, using the induction hypothesis, we get 7·(k + 1) = 7·(i + j) = 7·i + 7·j = 0 + 0 = 0. So P(k + 1) holds.

Solution

Identify that natural numbers N = { 1,2,3, --------} and not initial from n=0

the essential step must hold good for n=1 and not n=0

basic step is incorrect in the proof .

Basic step be P (1) = 7. 1= 7 \ eq0 . basic step is not applicable. therefore the statement that 7 .

n =0 is a incorrect statement