To prove that $\sqrt[3]{\sqrt{2}}$ is irrational, let's proceed with a proof by contradiction. The goal is to show that $\sqrt[3]{\sqrt{2}}$ cannot be expressed as a ratio of two integers.

Proof by Contradiction

1. Assumption: Suppose $\sqrt[3]{\sqrt{2}}$ is rational. This means that it can be written as a fraction of two integers:

   $\sqrt[3]{\sqrt{2}} = \frac{p}{q}$

   where $p$ and $q$ are integers, and the fraction is in its simplest form (i.e., $\gcd(p, q) = 1$).

2. Cubing Both Sides: Cube both sides of the equation to eliminate the cube root:

   $\sqrt{2} = \left(\frac{p}{q}\right)^3$

   Simplifying this, we get:

   $\sqrt{2} = \frac{p^3}{q^3}$

3. Squaring Both Sides: Square both sides to eliminate the square root:

   $2 = \left(\frac{p^3}{q^3}\right)^2$

   Simplify the right-hand side:

   $2 = \frac{p^6}{q^6}$

4. Rearrange the Equation: Multiply both sides by $q^6$:

   $2q^6 = p^6$

5. Implication on $p^6$: This equation implies that $p^6$ is even because it is equal to $2q^6$. Since $p^6$ is even, $p$ itself must also be even (as the power of an odd number is always odd).

   Let $p = 2k$ for some integer $k$.

6. Substitute $p = 2k$: Substitute back into the equation:

   $2q^6 = (2k)^6$

   Simplify this:

   $2q^6 = 64k^6$

   or

   $q^6 = 32k^6$

7. Implication on $q^6$: This implies that $q^6$ is also even, meaning $q$ is even.

8. Contradiction: Since both $p$ and $q$ are even, they have a common factor of 2, which contradicts our initial assumption that $\gcd(p, q) = 1$.

Conclusion

Since assuming $\sqrt[3]{\sqrt{2}}$ is rational led to a contradiction, it must be irrational.

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Would you like more details on this proof or have any questions? Here are some related questions:

1. How do you prove that $\sqrt{2}$ is irrational?
2. What is the general approach to proving a number is irrational?
3. Can other roots, like $\sqrt[4]{\sqrt{5}}$, be rational?
4. Why is a proof by contradiction often used for irrationality proofs?
5. How can you determine if other roots are irrational?

Tip: Proof by contradiction is a powerful method, especially when direct proof is challenging or indirect properties are more accessible.