求证：对任意正实数u及x<y，存在有理数r使x<ru<y，该证明是否充分？

Yes, this proof is fully rigorous—assuming the density of rationals in the real numbers is a result you can cite (which it almost always is in introductory analysis courses). Let’s break down why each step holds without gaps:

• Step 1: x/u < y/u
  Since u > 0, dividing both sides of x < y by u preserves the inequality direction (a basic property of real numbers for positive scalars). Both x/u and y/u are real numbers, so we’re working within the set ℝ where density applies.

• Step 2: Existence of r ∈ ℚ with x/u < r < y/u
  The density theorem for rationals states that between any two distinct real numbers, there exists at least one rational number. Since x/u < y/u, this theorem directly guarantees such an r exists.

• Step 3: Multiply by u to get x < ru < y
  Again, since u > 0, multiplying all parts of the inequality by u preserves the order. This transforms the rational r into ru, which satisfies the original problem’s requirement.

The only potential "gap" would be if you were expected to prove the density theorem itself (rather than using it as a lemma). But in most contexts, citing density is standard and acceptable—this proof doesn’t skip any necessary steps that aren’t already covered by established results.

内容的提问来源于stack exchange，提问作者Arjeta