Prove that cubed root of square root of 2 is irrational

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.

Explore a detailed proof that the cubed root of the square root of 2 is irrational using proof by contradiction. This problem delves into key concepts of number theory, including the properties of irrational numbers and even and odd integers. Ideal for students in Grades 10-12.

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