Question

Let x and y be two real numbers, then which of the following is always true ?

• A. $|\text{x} - \text{y}| \le |\text{x}| - |\text{y}|$
• B. $|\text{x} - \text{y}| \le |\text{x}| + |\text{y}|$
• C. $|\text{x} + \text{y}| \le |\text{x}| - |\text{y}|$
• D. $|\text{x} + \text{y}| \ge |\text{x}| + |\text{y}|$

Hint:

The absolute value of the sum or difference of two numbers is always less than or equal to the sum of their individual absolute values:
\[ |x \pm y| \le |x| + |y| \]
This is the most common form of the Triangle Inequality.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We utilize the absolute value triangle inequalities for real numbers.
Key Formula or Approach:
The standard triangle inequality states that for any real numbers \( a \) and \( b \):
\[ |a + b| \le |a| + |b| \]

Step 2: Detailed Explanation:
Let us test the inequalities using the standard triangle inequality and counterexamples:
- Evaluating Option B:
Let us rewrite \( x - y \) as \( x + (-y) \).
Applying the standard triangle inequality:
\[ |x - y| = |x + (-y)| \le |x| + |-y| \]
Since \( |-y| = |y| \) for any real number \( y \), we substitute this back into the inequality:
\[ |x - y| \le |x| + |y| \]
This inequality is always true for any real numbers \( x \) and \( y \), confirming Option B.
- Evaluating other options with counterexamples:
Let \( x = 2 \) and \( y = -3 \).
- For Option A: \( |2 - (-3)| = |5| = 5 \). However, \( |x| - |y| = 2 - 3 = -1 \). Since \( 5 \le -1 \) is false, Option A is not always true.
- For Option C: \( |2 + (-3)| = |-1| = 1 \). However, \( |x| - |y| = 2 - 3 = -1 \). Since \( 1 \le -1 \) is false, Option C is not always true.
- For Option D: \( |2 + (-3)| = 1 \). However, \( |x| + |y| = 2 + 3 = 5 \). Since \( 1 \ge 5 \) is false, Option D is not always true.

Step 3: Final Answer:
The inequality that is always true is $|\text{x} - \text{y}| \le |\text{x}| + |\text{y}|$.