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$$2+x$$

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Theorem

There exist irrational numbers $a$ and $b$ such that $a^b$ is rational.

Proof

We will prove this by constructing a specific example.

1. Let $a = \sqrt{2}$. We know $\sqrt{2}$ is irrational.

2. Consider $(\sqrt{2})^{\sqrt{2}}$. We have two possibilities:

   a) If $(\sqrt{2})^{\sqrt{2}}$ is rational, we've found our example and the proof is complete.

   b) If $(\sqrt{2})^{\sqrt{2}}$ is irrational, we proceed to the next step.

3. Assume $(\sqrt{2})^{\sqrt{2}}$ is irrational. Let's call this number $x$:

   $x = (\sqrt{2})^{\sqrt{2}}$

4. Now consider $x^{\sqrt{2}}$. We can rewrite this as:

   $x^{\sqrt{2}} = ((\sqrt{2})^{\sqrt{2}})^{\sqrt{2}}$

5. Using the laws of exponents, we can simplify:

   $((\sqrt{2})^{\sqrt{2}})^{\sqrt{2}} = (\sqrt{2})^{\sqrt{2} \cdot \sqrt{2}} = (\sqrt{2})^2 = 2$

6. 2 is clearly rational.

Therefore, we have shown that $x^{\sqrt{2}} = 2$, where $x$ is irrational (by our assumption in step 3) and $\sqrt{2}$ is irrational.

Conclusion

We have constructed irrational numbers $a$ and $b$ (namely, $x$ and $\sqrt{2}$) such that $a^b$ is rational (specifically, equal to 2).

This proves that it's possible for an irrational number raised to an irrational power to be rational.

Q.E.D.