1.

Prove that xn – yn is divisible by x – y; when n is a + ve integer.

Answer»

Let T(n) be the statement: 

xn – yn is divisible by x – y. 

Basic Step:

For n = 1, x1 – y1 = x – y is divisible by (x – y) 

⇒ T(1) is true 

Induction Step: 

Assume that T(k) is true, i.e., for k∈N xk – yk is divisible by (x – y) 

Now, we prove T(k + 1) is true. 

xk+1 – yk+1 = xk . x – yk . y = xk.x – xk . y + xk .y – yk . y 

(Adding and subtracting xk.y) 

= xk (x – y) + y (xk – yk ) Since xk(x – y) is divisible by (x – y) and (xk – yk ) is divisible by (x – y) (By induction step, i.e., assuming T(k) is true), therefore,

xk+1 – yk+1 = xk(x – y) + y(xk – yk) is divisible by (x – y) ⇒ T(k + 1) is true, whenever T(k) is true. 

⇒ T(n) holds for all positive integral values of n.


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