Consider the following in respect of the numbers `sqrt2,root(3)(2)androot(6)(6)`
I. `root(6)(6)` is the greatest number.
II. `sqrt2` is the smallest number.
Which of the above statements is/are correct?

Options

A. Only I
B. Only II
C. Both I and II
D. Neither I nor II

Answer

To determine which of the numbers \( \sqrt{2} \), \( \sqrt[3]{2} \), and \( \sqrt[6]{6} \) is the greatest and which is the smallest, we will follow these steps:

### Step 1: Convert the roots to exponential form
We can express each of the numbers in exponential form:
- \( \sqrt{2} = 2^{1/2} \)
- \( \sqrt[3]{2} = 2^{1/3} \)
- \( \sqrt[6]{6} = 6^{1/6} \)

### Step 2: Find a common denominator for the exponents
To compare these numbers easily, we can convert them to have a common denominator. The least common multiple (LCM) of the denominators (2, 3, and 6) is 6. We will rewrite each exponent:
- \( \sqrt{2} = 2^{1/2} = 2^{3/6} \)
- \( \sqrt[3]{2} = 2^{1/3} = 2^{2/6} \)
- \( \sqrt[6]{6} = 6^{1/6} \)

### Step 3: Rewrite \( \sqrt[6]{6} \) in terms of base 2
Next, we can express \( 6 \) in terms of its prime factors:
- \( 6 = 2 \times 3 \)
Thus, we can write:
- \( \sqrt[6]{6} = (2 \times 3)^{1/6} = 2^{1/6} \times 3^{1/6} \)

### Step 4: Compare the values
Now we have:
- \( \sqrt{2} = 2^{3/6} \)
- \( \sqrt[3]{2} = 2^{2/6} \)
- \( \sqrt[6]{6} = 2^{1/6} \times 3^{1/6} \)

To compare \( 2^{3/6} \), \( 2^{2/6} \), and \( 2^{1/6} \times 3^{1/6} \), we can evaluate the approximate numerical values:
- \( \sqrt{2} \approx 1.414 \)
- \( \sqrt[3]{2} \approx 1.260 \)
- \( \sqrt[6]{6} \approx 1.348 \)

### Step 5: Determine the greatest and smallest numbers
From the numerical approximations:
- \( \sqrt{2} \) is approximately \( 1.414 \) (greatest)
- \( \sqrt[6]{6} \) is approximately \( 1.348 \)
- \( \sqrt[3]{2} \) is approximately \( 1.260 \) (smallest)

### Conclusion
- The statement "root(6)(6) is the greatest number" is **incorrect**.
- The statement "sqrt2 is the smallest number" is **incorrect**.

Both statements are wrong.