Question

Let a and b be two real numbers with a, b $>$ 0 and a $<$ b, then which of the following is true

• A. $\sqrt{a} < \sqrt{b}$
• B. $\sqrt{b} < \sqrt{a}$
• C. $\text{b}^2 < \text{a}^2$
• D. $\text{a}^2 - \text{b}^2 > 0$

Hint:

For positive numbers, operations like squaring, cubing, or taking the square root always preserve the direction of the inequality, whereas taking the reciprocal reverses it (i.e., \( a \frac{1}{b} \)).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We examine the properties of inequalities for positive real numbers under monotonic mathematical operations such as taking square roots or squaring.
Key Formula or Approach:
For any positive real numbers \( a \) and \( b \):
If \( a < b \), then any strictly increasing function \( f(x) \) preserved the inequality order, i.e., \( f(a) < f(b) \).

Step 2: Detailed Explanation:
We are given that \( a, b > 0 \) and \( a < b \).
- Evaluating the square root function:
The function \( f(x) = \sqrt{x} \) is strictly increasing for all \( x > 0 \), because its derivative \( f'(x) = \frac{1}{2\sqrt{x}} > 0 \) is always positive.
Since the function is strictly increasing, applying it to both sides of the inequality \( a < b \) preserves the inequality direction:
\[ \sqrt{a} < \sqrt{b} \]
This shows that Option A is true, and Option B (\( \sqrt{b} < \sqrt{a} \)) is false.
- Evaluating the squaring function:
For positive numbers, the function \( g(x) = x^2 \) is also strictly increasing.
Applying it to \( a < b \) yields:
\[ a^2 < b^2 \]
This shows that Option C (\( b^2 < a^2 \)) is false.
Subtracting \( b^2 \) from both sides of \( a^2 < b^2 \) gives:
\[ a^2 - b^2 < 0 \]
This shows that Option D (\( a^2 - b^2 > 0 \)) is false.

Step 3: Final Answer:
The correct inequality is $\sqrt{a} < \sqrt{b}$.