If a and b are two distinct natural numbers, which one of the following is true?

Options

A. `sqrt(a+b) gt sqrt(b) + sqrt(b)`
B. `sqrt(a+b)= sqrt(a)+sqrt(b)`
C. `sqrt(a+b) lt sqrt(a)+ sqrt(b)`
D. `ab=1`

Answer

To solve the problem, we need to compare the two expressions involving distinct natural numbers \( a \) and \( b \):

1. **Expressions to Compare**:
- \( \sqrt{a + b} \)
- \( \sqrt{a} + \sqrt{b} \)

2. **Squaring Both Sides**:
To compare these two expressions, we will square both sides. This is valid because both expressions are non-negative (since \( a \) and \( b \) are natural numbers).

Let's denote:
- \( x = \sqrt{a + b} \)
- \( y = \sqrt{a} + \sqrt{b} \)

Now we square both sides:
\[ x^2 = a + b \]
\[ y^2 = (\sqrt{a} + \sqrt{b})^2 = a + b + 2\sqrt{ab} \]

3. **Comparing the Squared Values**:
Now we compare \( x^2 \) and \( y^2 \):
\[ x^2 = a + b \]
\[ y^2 = a + b + 2\sqrt{ab} \]

Here, we can see that:
\[ y^2 = x^2 + 2\sqrt{ab} \]

4. **Conclusion**:
Since \( \sqrt{ab} \) is a positive number (as \( a \) and \( b \) are distinct natural numbers), it follows that:
\[ y^2 > x^2 \]
Therefore, we conclude that:
\[ \sqrt{a} + \sqrt{b} > \sqrt{a + b} \]

5. **Final Result**:
The correct statement is:
\[ \sqrt{a} + \sqrt{b} > \sqrt{a + b} \]