written 8.0 years ago by | modified 2.9 years ago by |
Mumbai University > Computer Engineering > Sem 3 > Discrete Structures
Marks: 6 Marks
Year: Dec 2015
written 8.0 years ago by | modified 2.9 years ago by |
Mumbai University > Computer Engineering > Sem 3 > Discrete Structures
Marks: 6 Marks
Year: Dec 2015
written 8.0 years ago by |
Let the given statement be P (n). Then,
P(n): (ab)n = anbn.
When = 1, LHS = (ab)1 = ab and RHS = a1b1 = ab
Therefore LHS = RHS.
Thus, the given statement is true for n = 1, i.e., P(1) is true.
Let P(k) be true. Then,
P(k): (ab)k = akbk.
Now, (ab)k + 1 = (ab)k (ab)
= (akbk)(ab) [using (i)]
= (ak ∙ a)(bk ∙ b) [by commutativity and associativity of multiplication on real numbers]
= (ak + 1 ∙ bk + 1 ).
Therefore P(k+1): (ab)k + 1 = ((ak + 1 ∙ bk + 1)
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.