Question

Prove that the cube root of a prime number, p >or=2, is irrartional

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Santino

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To prove that the cube root of a prime number, p ≥ 2, is irrational, we can use proof by contradiction.

Assume that ∛p is rational. Then, there exist two coprime integers a and b (b ≠ 0) such that∛p = \frac{a}{b} , where a and b have no common factors other than 1.

Cubing both sides, we get:

p = \frac{a^3}{ b^3}

Since p is a prime number, it only has two distinct positive divisors: 1 and p. This means thata^3 is equal to p and b^3 is equal to 1, or vice versa.

Ifa^3 = p , then a=p^{\frac{1}{3}} , which contradicts the assumption that a is an integer. If b^3 = p , then b=p^{\frac{1}{3}} , which again contradicts the assumption that b is an integer.

Therefore, our assumption that ∛p is rational leads to a contradiction, and we conclude that the cube root of a prime number, p ≥ 2, is irrational.

\boxed{\text{Answer: The cube root of a prime number, } p \geq 2, \text{ is irrational.}}

Assume that ∛p is rational. Then, there exist two coprime integers a and b (b ≠ 0) such that

Cubing both sides, we get:

Since p is a prime number, it only has two distinct positive divisors: 1 and p. This means that

If

Therefore, our assumption that ∛p is rational leads to a contradiction, and we conclude that the cube root of a prime number, p ≥ 2, is irrational.

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