Question

The distributive law from algebra says that for all real numbers c, a₁ and a₂, 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, ..., a_n are any real numbers, then c(a₁ + a₂ + ... + a_n) = ca₁ + ca₂ + ... + ca_n.

Answer 1

Let P(n) be the given statement

i.e., P(n) : c(a₁ + a₂ + ... + a_n) = ca₁ + ca₂ + ... ca_n for all natural numbers n ≥ 2, for c, a₁, a₂, ... a_n ∈ R.

We observe that P(2) is true.

Since c(a₁ + a₂) = ca₁ + ca₂   ......(By distributive law)

Assume that P(n) is true for some natural number k

Where k > 2,

i.e., P(k) : c(a₁ + a₂ + ... + a_k) = ca₁ + ca₂ + ... + ca_k

Now to prove P(k + 1) is true.

We have P(k + 1): c(a₁ + a₂ + ... + a_k + a_{k + 1})

= c((a₁ + a₂ + ... + a_k) + a_{k + 1})

= c(a₁ + a₂ + ... + a_k) + ca_k+1    ........(By Distributive law)

= ca₁ + ca₂ + ... + ca_k + ca_{k + 1}

Thus P(k + 1) is true.

Whenever P(k) is true.

Hence, by the principle of Mathematical Induction, P(n) is true for all natural numbers n ≥ 2.