Question

Which of the following statements is false?

• A. If $a$ and $b$ are positive real numbers, then $\sqrt{ab} \le (a+b)/2$
• B. If $0 \le a < b$ ($a, b \in \mathbb{R}$), then $a^2 < ab < b^2$
• C. $\sup \{1 - 1/n : n \in \mathbb{N}\} = 1$
• D. If $a$ is a real number such that $0 < a < 1$, then $0 < a^2 < a < 1$

Hint:

Always test boundary values such as $a = 0$ when an inequality is defined with a weak inequality like $0 \le a$. This is a fast way to find counterexamples.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question tests basic properties of real numbers, inequalities, and bounds.

Step 2: Detailed Explanation:
Let us evaluate each statement:
- (A) "If $a$ and $b$ are positive real numbers, then $\sqrt{ab} \le (a+b)/2$": This is the classical Arithmetic Mean-Geometric Mean (AM-GM) inequality, which is always true.
- (C) "$\sup \{1 - 1/n : n \in \mathbb{N}\} = 1$": As $n$ increases, the terms $1 - 1/n$ approach 1 from below, but never exceed 1. Thus, the least upper bound (supremum) is indeed 1. This is true.
- (D) "If $0 < a < 1$, then $0 < a^2 < a < 1$": For a fraction between 0 and 1, squaring it makes it smaller (e.g., $0.5^2 = 0.25 < 0.5$). This is true.
- (B) "If $0 \le a < b$, then $a^2 < ab < b^2$": Let us test the boundary case where $a = 0$ and $b = 2$:
Since $0 \le 0 < 2$ is true, we substitute these into the inequality:
\[ 0^2 < 0 \cdot 2 < 2^2 \implies 0 < 0 < 4 \]
The inequality $0 < 0$ is false. The correct weak inequality should be $a^2 \le ab < b^2$.
Thus, statement (B) is false.

Step 3: Final Answer
The correct option is (B).