Chapter 5.1, Problem 35WE

To determine

Prove $\frac{{\mathrm{a}}^{\mathrm{m}}}{{\mathrm{a}}^{\mathrm{n}}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}};\mathrm{m}>\mathrm{n}$

Explanation of Solution

Given :

m and n are positive integers.

Calculation:

  $\frac{{\mathrm{a}}^{\mathrm{m}}}{{\mathrm{a}}^{\mathrm{n}}}=\frac{\underset{\mathrm{n}\text{ terms}}{\underbrace{\mathrm{a}\cdot \mathrm{a}\cdot \mathrm{a}\cdot \mathrm{...}\cdot \mathrm{a}}}\cdot \underset{(\mathrm{m}-\mathrm{n})\text{ terms}}{\underbrace{\mathrm{a}\cdot \mathrm{a}\cdot \mathrm{a}\cdot \mathrm{...}\cdot \mathrm{a}}}}{\underset{\mathrm{n}\text{ terms}}{\underbrace{\mathrm{a}\cdot \mathrm{a}\cdot \mathrm{a}\cdot \mathrm{...}\cdot \mathrm{a}}}}$

Now cancel same number of factors of a from the numerator and denominator .

Since m > n , the n factors of a in the numerator and denominator are cancelled.

Remaining are the ( m − n ) factors of a in the numerator .

Thus after cancellation , ${\mathrm{a}}^{\mathrm{m}-\mathrm{n}}$ is left in the numerator.

Hence,

  $\frac{{\mathrm{a}}^{\mathrm{m}}}{{\mathrm{a}}^{\mathrm{n}}}=\frac{\underset{\mathrm{m}}{\underbrace{\mathrm{a}\cdot \mathrm{a}\cdot \mathrm{a}\cdot \mathrm{...}\cdot \mathrm{a}}}}{\underset{\mathrm{n}}{\underbrace{\mathrm{a}\cdot \mathrm{a}\cdot \mathrm{a}\cdot \mathrm{...}\cdot \mathrm{a}}}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}}$ ; if m > n .

Hence proved.