Question

The distributive law from algebra states that for real numbers 

c, a₁ and a_2, we have c(a₁ + a₂) = ca₁ + ca₂ 

Use this law and mathematical induction to prove that, for all 

natural numbers, n ≥ 2, if c, a₁, a₂, …... an are any real numbers, 

then c(a₁ + a₂ +…+ an) = ca₁ + ca₂ +…+ ca_n.

Answer 1

Let P(n):c(a₁+a₂+…+an) = ca₁+ca₂+…+ca_n ,for all natural 

numbers, n ≥ 2. 

Step1: For n=2, 

P(2) 

LHS= c(a₁ + a₂) 

RHS= c a₁ + ca₂ 

As, it is given that c(a₁ + a₂) = c a₁ + ca₂ 

Thus, P(2) is true. 

Step2: For n=k, 

Let P(k) be true 

So, c(a₁+a₂+…+a_k ) = ca₁+ca₂+…+ca_k 

Now, we need to show P(k+1) is true whenever P(k) is true.

P(k+1): 

LHS= c(a₁+a₂+…+a_K+a_k+1) 

=c[(a₁+a₂+…+a_K)+a_k+1] 

=c(a₁+a₂+…+a_K)+ca_k+1 

=ca₁+ca₂+…+ca_K+ca_k+1 

= RHS 

Thus, P(k+1) is true, so by mathematical induction P(n) is true.