Question

For every natural number k, which of the following is true?

(a) (mn)^k = m^kn^k

(b) m*k = n + 1

(c) (m+n)^k = k + 1

(d) m^kn = mn^k

The question was asked in an interview for job.

Enquiry is from Principle of Mathematical Induction topic in portion Induction and Recursion of Discrete Mathematics

Answer 1

Right answer is (a) (mn)^k = m^kn^k

Explanation: In the first step, for k = 1, (mn)^1 = m^1n^1 = mn, hence it is true. Let us assume the statement is true for k = l, Now by induction assumption, (mn)^1 = m^1n^1 is true. So, to prove, (mn)^l+1 = m^l + 1n^l+1, we have (mn)^l = m^ln^l and multiplying both sides by (mn) ⇒ (mn)^1(mn)=(m^1n^1)(mn)

⇒ (mn)^l+1 = (mm^1)(nn^1) ⇒  (mn)^l+1 = (m^l+1n^l+1). Hence, it is proved.  So, (mn)^k = m^kn^k is true for every natural number k.