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Question
Prove the statement by using the Principle of Mathematical Induction:
For any natural number n, xn – yn is divisible by x – y, where x and y are any integers with x ≠ y.
Solution
P(n) = xn – yn is divisible by x – y, x integers with x ≠ y.
So, substituting different values for n, we get,
P(0) = x0 – y0 = 0 Which is divisible by x − y.
P(1) = x − y Which is divisible by x − y.
P(2) = x2 – y2
= (x + y)(x − y) Which is divisible by x − y.
P(3) = x3 – y3
= (x − y)(x2 + xy + y2) Which is divisible by x − y.
Let P(k) = xk – yk be divisible by x – y;
So, we get,
⇒ xk – yk = a(x − y).
Now, we also get that,
⇒ P(k + 1) = xk+1 – yk+1
= xk(x − y) + y(xk − yk)
= xk(x − y) + ya(x − y) Which is divisible by x − y.
⇒ P(k + 1) is true when P(k) is true.
Therefore, by Mathematical Induction,
P(n) xn – yn is divisible by x – y, where x integers with x ≠ y which is true for any natural number n.
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