Using Properties of Exponents to Derive Rational Exponents

Extending Properties From Integer to Rational Exponents

While extending the notion of exponents from integers to rationals, we want to make sure of few things.

1. We want the laws of exponents for integer exponents to be valid for rational exponents.

2. We want to introduce notion of nth root or inverse function of $f(x)=x^n$.

Lets try to define $a^{\frac{1}{n}}, \, n \in \mathbb{N}$. Consider $(a^{n})^{\frac{1}{n}}$. As we want the laws of exponents to hold, we want $(a^{n})^{\frac{1}{n}}=a^{n \times \frac{1}{n}} =a^1=a$.

So, we want to define $a^{\frac{1}{n}} = \sqrt[n]{a}$. In other words we want $a^{\frac{1}{n}} =b , b \in \mathbb{R}$ where $b$ is such that $b^n=a$. Lets try solving some equations of form $b^n=a$.

There is a way to get away from this problem. We restrict $a$ and also the value of $a^x ,x \in \mathbb{Q}$ to non negative real numbers. So now we want $\forall a\in \mathbb{R}_{\ge 0} a^{\frac{1}{n}} = b$ where $b\in\mathbb{R}_{\ge 0}$. But there is still a problem with defining $a^{\frac{1}{n}}$ as we want to.

So now we define $\forall a\in \mathbb{R}_{\ge 0}\,\, a^{\frac{1}{n}} = b$ where $b\in\mathbb{R}_{\ge 0}$ is such that $b^n=a$.

As we want laws of exponents for integers to hold, we must define $a^\frac{m}{n}=\left (a^{\frac{1}{n}} \right)^m$. These two definitions define the value of $a^x\, \forall a \in \mathbb{R}_{\ge 0}\, and \, \forall x \in \mathbb{Q}$.

Laws of exponents

Now we can prove the laws of exponents for rational exponents as stated below, proofs of which are quiet easy and follow from definitions and laws of integer exponents. Here $a,b\in\mathbb{R}_{\ge 0}$ and $x,y \in \mathbb{Q}$

1. $\left(a^x \right)^y=a^{xy}$

2. $a^{-x} = \dfrac{1}{a^x}$

3. $a^x\times a^y= a^{x+y}$

4. $a^x \times b^x= (a \times b)^x$

5. If $a < b$ and $x > 0$ then $a^x < b^x$

Thus we now have managed to extend integer exponents to rational exponents so that the laws of exponents still hold and we have a way to express roots of real numbers.

But, as a price of this extension, we must restict the definition to only half of the rational numbers.

Time for some Practice...

Existence and Uniqueness of $a^{\frac{1}{n}}$

$\forall a \in \mathbb{R}_{\ge 0} \exists ! \, b \in \mathbb{R}_{\ge 0}$ such that $b^n=a$

Let $X=\{y \in \mathbb{R}_{\ge 0},\, y^n \le a \}$ Note that $0 \in X$ and $x$ is bounded above either by $1$ or by $a$ itself. Let $b=\sup(X)$. We claim that $b^n=a$. Suppose not.

If $b^n < a$ then consider $b+\frac{1}{k}, \, k \in \mathbb{n}$ such that $k > \frac{2^n-1}{a-b^n} \, if \, b \le 1$ or $k > \frac{b^{n-1}(2^n-1)}{a-b^n} \, if \, b \ge 1$.

If $b \le 1$ then $$\begin{align*} \left ( b + \frac{1}{k} \right )^n & = b^n + \frac{\binom{n}{1} b^{n-1}}{k^1} + \frac{\binom{n}{2} b^{n-2}}{k^2} + \cdots + \frac{\binom{n}{n} b^{0}}{k^n}\\ & < b^n + \frac{2^n-1}{k}\\ & < b^n + a - b^n = a \end{align*}$$ Thus $b + \frac{1}{k} \in X \Rightarrow b \neq \sup(X)$

If $b \ge 1$ then $$\begin{align*} \left ( b + \frac{1}{k} \right )^n & = b^n + \frac{\binom{n}{1} b^{n-1}}{k^1} + \frac{\binom{n}{2} b^{n-2}}{k^2} + \cdots + \frac{\binom{n}{n} b^{0}}{k^n}\\ & < b^n + \frac{b^{n-1}(2^n-1)}{k}\\ & < b^n + a - b^n = a \end{align*}$$ Thus $b + \frac{1}{k} \in X \Rightarrow b \neq \sup(X)$

Similarly, if $b^n > a$ then we can take $b - \frac{1}{k}$ where $k$ is defined as above to guarantee that $\left ( b - \frac{1}{k} \right)^n > a \Rightarrow b \neq \sup(x)$

Suppose there is $b$ such that $b^n=a$ then $b=\sup(X)$. Hence, such $b$ exists and is unique.

(Video) Rational Exponents

by yaymath

YayMath has a great video explaining the properties of Rational Exponents.