关于实数区间上全称量词命题的证明方法问询

Hey Hayden, no stress at all—this type of universal quantifier proof is totally manageable once you recall the core idea! Let’s walk through this problem step by step, and I’ll explain the general approach so you can apply it to similar problems later.

First, let’s recap what the universal quantifier ($\forall$) is asking us to do: we need to show that every single real number x that falls in the interval [1, 2] will make 3x - 1 fall in [2, 5]. The key trick for $\forall$ proofs is this: instead of trying to check every possible x (which is impossible, since there are infinitely many!), we just pick an arbitrary x that satisfies the premise (x ∈ [1,2]), then prove the conclusion holds for this x. Since our x was chosen arbitrarily, the result applies to all such x.

Let’s apply that to your problem:

• Start by stating: Fix an arbitrary real number x where 1 ≤ x ≤ 2 (this is just restating x ∈ [1,2] using the definition of closed intervals).
• Manipulate the inequality to derive the conclusion about 3x - 1:
  • Multiply every part of the inequality $1 ≤ x ≤ 2$ by 3 (since 3 is a positive number, the inequality signs stay the same):
    $3 \times 1 ≤ 3x ≤ 3 \times 2$ simplifies to $3 ≤ 3x ≤ 6$
  • Subtract 1 from every part of the new inequality:
    $3 - 1 ≤ 3x - 1 ≤ 6 - 1$ simplifies to $2 ≤ 3x - 1 ≤ 5$
• By the definition of the closed interval [2,5], the inequality $2 ≤ 3x - 1 ≤ 5$ directly means $(3x - 1) ∈ [2,5]$.

Since we didn’t make any special assumptions about x—we just picked it as any number in [1,2]—this conclusion holds for all real numbers x that meet the premise. That’s exactly what the original $\forall$ statement is claiming!

A quick pro tip for future $\forall$ proofs: always lead with picking an arbitrary element that fits the premise, then use logical or algebraic steps to work your way to the conclusion. Keep it straightforward, and you’ll nail every one of these.

备注：内容来源于stack exchange，提问作者Hayden