请求证明素数的k次根（k≥2，k∈ℕ）不属于有理数集

Hey there, I totally get how frustrating it is when your lecturer skips key explanations and even messes up solutions—let’s work through this proof step by step, no confusing gaps left behind.

First, let’s get aligned on key definitions and fix the direction of your approach:

• Rational numbers can always be written as a fraction a/b where a and b are integers, b ≠ 0, and gcd(a, b) = 1 (this means the fraction is in simplest form—no shared factors besides 1).
• The part you worked on about a + b ∈ ℚ is about rational numbers being closed under addition, which doesn’t apply here. We’ll use proof by contradiction and the fundamental properties of prime numbers instead.

Here’s the full, clear proof:

1. Start with a contradictory assumption: Suppose there does exist a rational number q such that q^k = p. By the definition of rational numbers, we can write q = a/b where a,b ∈ ℤ, b ≠ 0, and gcd(a,b) = 1.

2. Substitute into the equation:

   (a/b)^k = p
   

   Multiply both sides by b^k to eliminate the denominator:

   a^k = p * b^k
   

3. Analyze divisibility by prime p:

   • Since p is prime and divides the right-hand side (p*b^k), it must divide the left-hand side a^k. A core property of primes says: if a prime divides a product of integers, it divides at least one of the integers. Here, a^k is a multiplied by itself k times, so p must divide a.

4. Rewrite a with the factor of p:
   Let a = p*m where m ∈ ℤ. Substitute this back into the equation:

   (p*m)^k = p*b^k
   p^k * m^k = p*b^k
   

   Divide both sides by p (since p is prime, it’s non-zero):

   p^(k-1) * m^k = b^k
   

5. Spot the contradiction:

   • Now, p^(k-1) divides the left-hand side, so it must divide b^k. Since k ≥ 2, k-1 ≥ 1—meaning p is a factor of b^k. Again using the prime property, p must divide b.

   • But wait! We assumed gcd(a,b) = 1 (no shared factors), but now we’ve shown p divides both a and b—this is a direct contradiction.

6. Final conclusion: Our initial assumption (that q ∈ ℚ exists with q^k = p) is false. Therefore, there is no such rational number q.

To tie this to your example: when k=2, this proves that √p (the square root of a prime) is irrational—just a specific case of this general proof.

备注：内容来源于stack exchange，提问作者ADinar