# Proof of Quotient Rule of Exponents with same base

$b$ is a literal. It is multiplied by itself $m$ times to form an exponential term $b^{\displaystyle m}$ and also multiplied by itself $n$ number of times to form another exponential term $b^{\displaystyle n}$ mathematically.

$(1) \,\,\,$ $b^{\displaystyle m}$ $\,=\,$ $\underbrace{b \times b \times b \times \ldots \times b}_{\displaystyle m \, factors}$

$(2) \,\,\,$ $b^{\displaystyle n}$ $\,=\,$ $\underbrace{b \times b \times b \times \ldots \times b}_{\displaystyle n \, factors}$

Now, the division rule of indices with same base can be derived by dividing the exponential term $b^{\displaystyle m}$ by the exponential term $b^{\displaystyle n}$.

$\dfrac{b^{\displaystyle m}}{b^{\displaystyle n}}$ $\,=\,$ $\dfrac{b \times b \times b \times \ldots \times b}{b \times b \times b \times \ldots \times b}$

Now, we have to consider two cases to find the quotient of the exponents with same base.

### m > n

Take, the exponent of exponential term $b^{\displaystyle m}$ in the numerator is greater than the exponent of the exponential term $b^{\displaystyle n}$ in the denominator.

$\implies$ $\require{cancel} \dfrac{b^{\displaystyle m}}{b^{\displaystyle n}}$ $\,=\,$ $\dfrac{\cancel{b} \times \cancel{b} \times \cancel{b} \times \ldots \times b}{\cancel{b} \times \cancel{b} \times \cancel{b} \times \ldots \times \cancel{b}}$

Each factor in the denominator is cancelled by the each factor in the denominator but there are some factors remain in the numerator in this case. The remaining total number of factors is equal to the total factors in denominator subtracted from total factors in the numerator.

$\implies$ $\dfrac{b^{\displaystyle m}}{b^{\displaystyle n}}$ $\,=\,$ $\underbrace{b \times b \times b \times \ldots \times b}_{\displaystyle m-n \, factors}$

Now, express the product of factors in exponential notation.

$\therefore \,\,\,\,\,\,$ $\dfrac{b^{\displaystyle m}}{b^{\displaystyle n}}$ $\,=\,$ $b^{\displaystyle m-n}$

### m < n

Assume, the total number of factors in the numerator is less than the total number of factors in the denominator.

$\dfrac{b^{\displaystyle m}}{b^{\displaystyle n}} \,=\, \dfrac{b \times b \times b \times \ldots \times b}{b \times b \times b \times \ldots \times b}$

$\implies \require{cancel} \dfrac{b^{\displaystyle m}}{b^{\displaystyle n}} \,=\, \dfrac{\cancel{b} \times \cancel{b} \times \cancel{b} \times \ldots \times \cancel{b}}{\cancel{b} \times \cancel{b} \times \cancel{b} \times \ldots \times b}$

In this case, there are some factors remain in the denominator and it is equal to the total factors in the denominator subtracted from the total factors in the numerator.

$\implies$ $\dfrac{b^{\displaystyle m}}{b^{\displaystyle n}}$ $\,=\,$ $\dfrac{1}{\underbrace{b \times b \times b \times \ldots \times b}_{\displaystyle n-m \, factors}}$

Now, write the product of the factors in exponential form.

$\therefore \,\,\,\,\,\,$ $\dfrac{b^{\displaystyle m}}{b^{\displaystyle n}}$ $\,=\,$ $\dfrac{1}{b^{\displaystyle n-m}}$

Therefore, it is proved that the quotient for the division of exponents with same base is equal to the difference of the exponents with same base.

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