Let A = `{3,pi,sqrt2,-5,3+sqrt7,2//7}`. The subset of A containing all the elements from it which are irrational numbers is

Options

A. `{pi,sqrt2,3+sqrt7}`
B. `{3,pi,2//7,-5,3+sqrt7}`
C. `{3,2//7,-5}`
D. `{3,-5}`

Answer

To find the subset of set A that contains all the irrational numbers, we will follow these steps:

### Step 1: Identify the elements of set A
Set A is given as:
\[ A = \{ 3, \pi, \sqrt{2}, -5, 3 + \sqrt{7}, \frac{2}{7} \} \]

### Step 2: Determine which elements are irrational
We need to classify each element of set A as either rational or irrational:
- **3**: Rational (can be expressed as \(\frac{3}{1}\))
- **\(\pi\)**: Irrational (cannot be expressed as a fraction)
- **\(\sqrt{2}\)**: Irrational (cannot be expressed as a fraction)
- **-5**: Rational (can be expressed as \(\frac{-5}{1}\))
- **\(3 + \sqrt{7}\)**: Irrational (since \(\sqrt{7}\) is irrational, the sum is also irrational)
- **\(\frac{2}{7}\)**: Rational (can be expressed as a fraction)

### Step 3: List the irrational numbers
From the classification above, the irrational numbers in set A are:
- \(\pi\)
- \(\sqrt{2}\)
- \(3 + \sqrt{7}\)

### Step 4: Form the subset of irrational numbers
The subset of A containing all the irrational numbers is:
\[ \{ \pi, \sqrt{2}, 3 + \sqrt{7} \} \]

### Final Answer
The subset of A containing all the elements from it which are irrational numbers is:
\[ \{ \pi, \sqrt{2}, 3 + \sqrt{7} \} \]

---