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^kThe question was asked in an interview for job.Enquiry is from Principle of Mathematical Induction topic in portion Induction and Recursion of Discrete Mathematics

For every natural number k, which of the following is true?(a) (mn)^k

1.	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^kThe 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» 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.

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