Prove by the principle of mathematical induction:(ab)n = an bn for

1.	Prove by the principle of mathematical induction:(ab)n = an bn for all n ϵ N
Answer» Suppose P (n): (ab)ⁿ = aⁿ bⁿ Now let us check for n = 1, P (1): (ab)¹ = a¹ b¹ : ab = ab P (n) is true for n = 1. Then, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true. P (k): (ab)^k = a^k b^k … (i) Now we have to prove, (ab)^{k + 1}= a^{k + 1}.b^{k + 1} Therefore, = (ab)^{k + 1} = (ab)^k (ab) = (a^kb^k) (ab) using equation (1) = (a^{k + 1}) (b^{k + 1}) P (n) is true for n = k + 1 Thus, P (n) is true for all n ∈ N.

Prove by the principle of mathematical induction:(ab)n = an bn for all n ϵ N

Answer»

Suppose P (n): (ab)ⁿ = aⁿ bⁿ

Now let us check for n = 1,

P (1): (ab)¹ = a¹ b¹

: ab = ab

P (n) is true for n = 1.

Then, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.

P (k): (ab)^k = a^k b^k … (i)

Now we have to prove,

(ab)^{k + 1}= a^{k + 1}.b^{k + 1}

Therefore,

= (ab)^{k + 1}

= (ab)^k (ab)

= (a^kb^k) (ab) using equation (1)

= (a^{k + 1}) (b^{k + 1})

P (n) is true for n = k + 1

Thus, P (n) is true for all n ∈ N.