Power with Real Exponent
Grade 10 · Algebra
1. Extending the Definition to Irrational Exponents
You already know how to compute $a^r$ when $r$ is a rational number (a fraction $\frac{p}{q}$). But what does $a^x$ mean when $x$ is irrational, like $\sqrt{2}$ or $\pi$?
Key Idea: Every irrational number can be approximated by a sequence of rational numbers. We use these approximations to define the power.
Example: Defining $3^{\sqrt{2}}$
The number $\sqrt{2} = 1.41421356\ldots$ is irrational. We build rational approximations:
| Step $n$ | Rational approx. $r_n$ | $3^{r_n}$ |
|---|---|---|
| 1 | 1.4 | $3^{1.4} \approx 4.6555$ |
| 2 | 1.41 | $3^{1.41} \approx 4.7070$ |
| 3 | 1.414 | $3^{1.414} \approx 4.7277$ |
| 4 | 1.4142 | $3^{1.4142} \approx 4.7287$ |
| 5 | 1.41421 | $3^{1.41421} \approx 4.7288$ |
Definition: $$a^x = \lim_{n \to \infty} a^{r_n}$$ where $r_1, r_2, r_3, \ldots$ are rational numbers approaching $x$.
So $3^{\sqrt{2}} \approx 4.7288\ldots$ — a well-defined positive real number.
Interactive: Watch $3^{r_n}$ converge to $3^{\sqrt{2}}$
Interactive visualization — available in the full lesson
Interactive visualization — available in the full lesson
2. General Definition
Definition: For any real number $x$ and any $a > 0$, the power $a^x$ is a positive real number.
$$a > 0 \implies a^x > 0 \quad \text{for all } x \in \mathbb{R}$$
Special Case: $a = 0$
| Condition | Value | Example |
|---|---|---|
| $x > 0$ | $0^x = 0$ ✅ | $0^5 = 0$, $0^{\sqrt{3}} = 0$ |
| $x = 0$ | $0^0$ is undefined ❌ | — |
| $x < 0$ | $0^x$ is undefined ❌ | $0^{-1}$, $0^{-\sqrt{2}}$ have no meaning |
⚠ Why is $0^{-1}$ undefined?
$0^{-1} = \frac{1}{0^1} = \frac{1}{0}$, which is division by zero — undefined!
⚠ Why is $0^{-\sqrt{2}}$ undefined?
$0^{-\sqrt{2}} = \frac{1}{0^{\sqrt{2}}} = \frac{1}{0}$ — again division by zero.
Remember: For $x \leq 0$, the expression $0^x$ has no meaning.