Let p(n) be the statement : " the arithmetic mean of n and (n+2) is same as their geometric mean ." PROVE THAT p(1) is NOT true. Also prove that if p(n) is true , then p(n+1) is also true. How does this contradict the principle of induction ?
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p(n) is the predicate : the A.M. = G.M. for n and n+2.
p(1) => the numbers are 1 and 3. AM = 2. GM =√(1*3) = √3
as AM ≠ GM , p(1) is false.
If p(n) is true, then (n+1)² = n (n +2).
now, for p(n+1) : AM of n+1, and n+3 is n+2 : so
(n+2)² = [(n+1)+1]²
= (n+1)² + 2 (n+1) + 1
= n (n+2) + 2 (n+1) + 1
= n² + 4 n + 3
= (n +1) (n +1+ 2)
=> p(n+1) is true.
p(n) : (n+1)² = n (n+2)
=> 1 = 0 => p(n) is not true for any n.
So the principle of mathematical induction is not applicable for p(n).
The principle of mathematical induction is not contradicted. If only p(n) is true for any n like n =1, then only p(n) is true for all n. As p(n) is not true for any n including n = 1, the predicate p(n) is false for all n.
p(1) => the numbers are 1 and 3. AM = 2. GM =√(1*3) = √3
as AM ≠ GM , p(1) is false.
If p(n) is true, then (n+1)² = n (n +2).
now, for p(n+1) : AM of n+1, and n+3 is n+2 : so
(n+2)² = [(n+1)+1]²
= (n+1)² + 2 (n+1) + 1
= n (n+2) + 2 (n+1) + 1
= n² + 4 n + 3
= (n +1) (n +1+ 2)
=> p(n+1) is true.
p(n) : (n+1)² = n (n+2)
=> 1 = 0 => p(n) is not true for any n.
So the principle of mathematical induction is not applicable for p(n).
The principle of mathematical induction is not contradicted. If only p(n) is true for any n like n =1, then only p(n) is true for all n. As p(n) is not true for any n including n = 1, the predicate p(n) is false for all n.
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