Prerequisites:
Exponentiation (definition)
Continuity of exponential function
Proof:
Case I: when x, y \in\mathbb{Z^+}
\qquad\quad \begin{align} a^{x+y} &= a.a.a.a......a\quad \text{($x+y$ times)}\\ &= \underbrace{(a.a.a.....a)}_\text{$x \text{ times}$}.\underbrace{(a.a.a.....a)}_\text{$y\text{ times}$}\\ &= a^x.a^y\end{align}
Case II: when x, y \in\mathbb{Z^-}
Let x_1 = -x, \;\; y_1 = -y and 1/a = b
\qquad\quad \begin{align} a^{x+y} &= a^{-(x_1+y_1)}\\\\ &= \left(\dfrac{1}{a}\right)^{(x_1+y_1)}\\\\ &= b^{(x_1+y_1)}\end{align}
Hence the case is reduced to the previous case and can be proved in a similar manner.
Case III: when x \in\mathbb{Z^+} and y \in\mathbb{Z^-}
Let y_1 = -y
\qquad\quad \begin{align} a^{x+y} &= a^{x-y_1}\\ &= a.a.a....a\quad\text{($x-y_1$ times)}\\\ &= \dfrac{a.a.a....a}{a.a.a....a}\begin{matrix}\text{($x$ times)}\\ \text{($y_1$ times)}\end{matrix}\\\ &= \dfrac{a^x}{a^{y_1}}\\\ &= a^x.a^{-y_1}\\ &= a^x.a^y\end{align}
Case IV: When atleast one of x, y equal to 0
Let y = 0
\qquad\quad \begin{align}a^0 &= a^{n-n} \qquad\quad\;\;\;\text{ where $n\in\mathbb{Z^+}$}\\ &= a^n.a^{-n}\qquad\quad\text{(from case III)}\\ &= \dfrac{a^n}{a^n} = 1\end{align}
Case V: When x, y \in\mathbb{Q}, i.e., x, y are rational numbers
As a corollary of case I, for c \in\mathbb{Z^+}
\qquad\quad\begin{align}\big(a^{b}\big)^c &= a^b.a^b.a^b....a^b\qquad\text{($c$ times)}\\ &= a^{(b+b+b...b)}\\ &= a^{\displaystyle (bc)}\end{align}
Let x = p_1/q_1 and y = p_2/q_2 where p_1, p_2, q_1, q_2 \in\mathbb{Z}\;\text{ and }\; q_1, q_2\neq 0
Let q' = q_1q_2,
\begin{align} \text{Now, }\quad a^{x+y} &= a^{\tfrac{p_1}{q_1}+ \tfrac{p_2}{q_2}}\\\\ &= a^{\tfrac{p_1q_2 + p_2q_1}{q'}}\\\\ &= \Big( a^{(1/q')}\Big) ^{(p_1q_2 + p_2q_1)}\qquad\qquad\qquad\text{(from the above corollary)}\\\\ &= \Big( a^{(1/q')}\Big) ^{(p_1q_2)}\Big( a^{(1/q')}\Big) ^{(p_2q_1)}\qquad\text{(from case I)}\\\\ &= a^{\left( \tfrac{p_1q_2}{q'}\right)}a^{\left( \tfrac{p_2q_1}{q'}\right)}\\\\ &= a^{\tfrac{p_1}{q_1}}.a^{\tfrac{p_2}{q_2}}\\\\ &= a^x.a^y\end{align}
Case VI: When x, y\in\mathbb{R}
If x and y are irrational, then in any neighbourhood of x and y, there exists infinitely many rational numbers. Since at these rational numbers, the given equality holds, hence by the continuity of the exponential functions, the equality must also hold at x and y.
Therefore the result holds at all real values of x and y.
Recommended:
Product of the powers
Power of the product
Log of product
Log of powers
No comments:
Post a Comment