涉及三角不等式的证明：两类实数不等式证明技术问询

Hey there, let's work through these two proofs step by step—they're straightforward once you leverage the standard triangle inequality and basic real number properties.

Part (a): If |x - y| < c, then |x| < |y| + c #

Start with the standard triangle inequality, a fundamental property of real numbers:

For any real numbers (a) and (b), (|a + b| \leq |a| + |b|)

We can rewrite (x) as a sum involving (y) and the difference (x - y):
x = y + (x - y)

Apply the triangle inequality to this sum:
|x| = |y + (x - y)| ≤ |y| + |x - y|

Now, substitute the given condition (|x - y| < c) into the right-hand side. Since inequalities are transitive, we get:
|x| ≤ |y| + |x - y| < |y| + c

Combining these gives exactly the result we need to prove:
|x| < |y| + c

Part (b): If for all (a > 0), (|x - y| < a), then (x = y) #

This is a classic proof by contradiction. Let's start by assuming the opposite of what we want to show: suppose (x \neq y).

If (x \neq y), then the absolute difference (|x - y|) is a positive real number. Let's call this value (d = |x - y|), where (d > 0).

Now, pick a positive (a) that's smaller than (d)—for example, let (a = \frac{d}{2}) (which is positive because (d > 0)).

By our assumption, (|x - y| = d), but (d > \frac{d}{2} = a). This directly contradicts the given condition that every positive (a) satisfies (|x - y| < a).

Since our assumption leads to a contradiction, it must be false. Therefore, (x = y).

内容的提问来源于stack exchange，提问作者N. Schuler