Does what I said make sense in this context "No rational number c satisfies c 3 = 2. Therefore, 3 square root 2 is irrational. Assume 3 square root 2 = $$ \frac{p}{q} $$. We'd have to then divide the equation by 3 so $$ \sqrt2 $$ = $$ \frac{p}{3q} $$However, this means that square root of 2 is a rational number which is a contradiction. It is well known that $$ \sqrt{2\text{ }} $$is irrational , therefore 3 square root of 2 must be irrational

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.

Does what I said make sense in this context: 'No rational number c satisfies c^3 = 2. Therefore, 3√2 is irrational. Assume 3√2 = p/q. We'd have to then divide the equation by 3 so √2 = p/3q. However, this means that square root of 2 is a rational number which is a contradiction. It is well known that √2 is irrational, therefore 3√2 must be irrational.'

Explore a detailed proof of the irrationality of 3√2 using the method of contradiction. Understand key algebraic concepts and theorems, including the irrationality of √2 and how assumptions can lead to contradictions. This problem is suitable for students in Grades 9-12.

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