An Irrational Power That Is Rational

In the HSC Mathematics Extension 2 course, the topic of proofs challenges students to move beyond calculation and into rigorous logical reasoning. This classic problem asks you to demonstrate that an irrational number raised to an irrational power can yield a rational result, a surprising fact that is proven by a clever case analysis on $\sqrt{2}^{\sqrt{2}}$. Through this proof, you will learn how to apply the Law of Excluded Middle to construct a non‑constructive existence argument—one that guarantees an example without ever having to identify which case actually occurs.

Problem Statement

Prove that an irrational number raised to the power of an irrational number can be rational, by considering $\sqrt{2}^{\sqrt{2}}$. You may assume $\sqrt{2}$ is irrational.

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Hints

Consider $\alpha = \sqrt{2}^{\sqrt{2}}$. By the Law of Excluded Middle, $\alpha$ is either rational or irrational. Examine both cases:

• If $\alpha$ is rational, you're done immediately.
• If $\alpha$ is irrational, compute $\alpha^{\sqrt{2}}$ and simplify.

This is a non-constructive existence proof---you prove something exists without determining which case actually holds!

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Solutions

Proof by Cases:

We begin by setting $\alpha = \sqrt{2}^{\sqrt{2}}$. By the Law of Excluded Middle, $\alpha$ must be either rational or irrational. We’ll show that in either scenario we can produce an example of the form (irrational)$^\text{(irrational)}$ that equals a rational number.

Case 1: If $\alpha = \sqrt{2}^{\sqrt{2}}$ is rational, then we have found irrational base $\sqrt{2}$ and irrational exponent $\sqrt{2}$ such that $\sqrt{2}^{\sqrt{2}}$ is rational. Done.

Case 2: Now assume $\alpha = \sqrt{2}^{\sqrt{2}}$ is irrational. We investigate $\alpha$ raised to the power $\sqrt{2}$. Using the index law $(a^m)^n = a^{mn}$, we simplify:

$$\begin{aligned} \alpha^{\sqrt{2}} &= \left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}} \\ &= \sqrt{2}^{\sqrt{2} \cdot \sqrt{2}} \quad \text{(exponent law)} \\ &= \sqrt{2}^{2} \\ &= 2 \end{aligned}$$

Since $2$ is rational and both $\alpha$ (irrational by case assumption) and $\sqrt{2}$ (given as irrational) are irrational, we have found an example.

Conclusion: In either case, there exist irrational numbers $a$ and $b$ such that $a^b$ is rational. $\blacksquare$

Note: This proof doesn't tell us whether $\sqrt{2}^{\sqrt{2}}$ is actually rational or irrational---and we don't need to know! This is the beauty of non-constructive existence proofs.

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Takeaways

• This is a non‑constructive existence proof: it uses the Law of Excluded Middle to split into two exhaustive cases without needing to know which case actually holds.
• Reconstruct the full proof from the hint and the solution outline, and justify every transformation explicitly. Check edge cases and verify where each assumption (like the irrationality of $\sqrt{2}$) is used in the argument.

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Further Readings

If you found this proof interesting, be sure to check out these relevant HSC booklets to sharpen your reasoning skills: HSC Proofs, HSC Trigonometry, HSC Differential Equations