Your reasoning is generally sound, but the presentation has a few areas where clarity could be improved. Here's a clearer and more mathematically precise version of your argument:

1. Initial Claim: "No rational number $c$ satisfies $c^3 = 2$. Therefore, $\sqrt[3]{2}$ is irrational."

   • This statement is correct. Since no rational number $c$ can be raised to the third power to equal 2, $\sqrt[3]{2}$ must be irrational.

2. Assumption for Contradiction: "Assume $3\sqrt[3]{2}$ is rational, i.e., $3\sqrt[3]{2} = \frac{p}{q}$ where $p$ and $q$ are integers with $q \neq 0$."

3. Derivation: "We then divide the equation by 3, giving us $\sqrt[3]{2} = \frac{p}{3q}$."

   • Here, you correctly derive that $\sqrt[3]{2}$ would be equal to a rational number $\frac{p}{3q}$ under the assumption that $3\sqrt[3]{2}$ is rational.

4. Contradiction: "However, this implies that $\sqrt[3]{2}$ is rational, which contradicts the well-known fact that $\sqrt[3]{2}$ is irrational."

5. Conclusion: "Therefore, $3\sqrt[3]{2}$ must be irrational."

Improvements:

• Notation: Ensure that the mathematical symbols are clearly defined and correctly notated. For instance, "3 square root 2" should be consistently written as $3\sqrt[3]{2}$ or "3 times the cube root of 2."
• Clarity: Explicitly state that your initial assumption leads to a contradiction, which is the basis for concluding that $3\sqrt[3]{2}$ is irrational.

Summary of the Revised Argument:

"No rational number $c$ satisfies $c^3 = 2$. Therefore, $\sqrt[3]{2}$ is irrational. Assume, for contradiction, that $3\sqrt[3]{2} = \frac{p}{q}$ where $p$ and $q$ are integers with $q \neq 0$. Dividing both sides by 3, we obtain $\sqrt[3]{2} = \frac{p}{3q}$, which would imply that $\sqrt[3]{2}$ is rational. This contradicts the fact that $\sqrt[3]{2}$ is irrational. Hence, $3\sqrt[3]{2}$ must be irrational."

Would you like more details or have any other questions?

Related Questions:

1. What is the general method to prove that a number is irrational?
2. Can you provide a similar proof for why $\sqrt{2}$ is irrational?
3. How does this reasoning extend to prove that $n\sqrt[m]{k}$ is irrational for integers $n$, $m$, and $k$?
4. Can irrational numbers be represented as a fraction?
5. What are some common misconceptions about rational and irrational numbers?

Tip:

When proving irrationality, always ensure that the proof involves assuming the number is rational and then showing this leads to a contradiction.