Question

Arrange the following in descending order.
(A) \(\sqrt{3} - \sqrt{2}\)
(B) \(\sqrt{4} - \sqrt{3}\)
(C) \(\sqrt{5} - \sqrt{4}\)
(D) \(\sqrt{2} - 1\)
Choose the correct answer from the options given below:

• A. \(A>B>C>D\)
• B. \(D>C>B>A\)
• C. \(D>A>B>C\)
• D. \(A>C>B>D\)

Hint:

For expressions like \(\sqrt{n+1} - \sqrt{n}\), the value decreases as \(n\) increases. You can quickly verify this with small numbers: \(\sqrt{2}-1 \approx 0.414\), while \(\sqrt{101}-10 \approx 0.05\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
To compare differences of square roots where the differences of the numbers inside are constant, we can rationalize the numerator to see the relationship more clearly.
Step 2: Key Formula or Approach:
\[ \sqrt{x+1} - \sqrt{x} = \frac{(\sqrt{x+1} - \sqrt{x})(\sqrt{x+1} + \sqrt{x})}{\sqrt{x+1} + \sqrt{x}} = \frac{1}{\sqrt{x+1} + \sqrt{x}} \]
Step 3: Detailed Explanation:
Using the rationalization logic for each term:
(A) \(\sqrt{3} - \sqrt{2} = \frac{1}{\sqrt{3} + \sqrt{2}}\)
(B) \(\sqrt{4} - \sqrt{3} = \frac{1}{\sqrt{4} + \sqrt{3}}\)
(C) \(\sqrt{5} - \sqrt{4} = \frac{1}{\sqrt{5} + \sqrt{4}}\)
(D) \(\sqrt{2} - 1 = \sqrt{2} - \sqrt{1} = \frac{1}{\sqrt{2} + \sqrt{1}}\)
Now compare the denominators. Since all numerators are \(1\), the term with the smallest denominator is the largest value.
Denominators comparison: \((\sqrt{2} + \sqrt{1})<(\sqrt{3} + \sqrt{2})<(\sqrt{4} + \sqrt{3})<(\sqrt{5} + \sqrt{4})\).
Therefore, the values in descending order are:
\( \frac{1}{\sqrt{2} + \sqrt{1}}>\frac{1}{\sqrt{3} + \sqrt{2}}>\frac{1}{\sqrt{4} + \sqrt{3}}>\frac{1}{\sqrt{5} + \sqrt{4}} \).
This corresponds to \(D>A>B>C\).
Step 4: Final Answer:
The descending order is \(D>A>B>C\).