Consider the set A = {x ∈ ℚ : 0 < (√2 - 1)x < √2 + 1} as a subset of ℝ. Which of the following statements is true?

Concept:

The supremum \(\sup A \) of the set is the least upper bound for \(x \) . The infimum \(\inf A \) is the greatest lower bound of \(x \)

We are given the set \( A = \{ x \in \mathbb{Q} : 0 < (\sqrt{2} - 1)x < \sqrt{2} + 1 \)} as a subset of \( \mathbb{R} \), and we need to determine

the correct supremum (sup) and infimum (inf) of this set.

We are asked to solve for  \(x\)  in the inequality \(0 < (\sqrt{2} - 1)x < \sqrt{2} + 1 \).

\(0 < (\sqrt{2} - 1)x < \sqrt{2} + 1\)

⇒ \(0 < x < \frac{\sqrt{2} + 1}{\sqrt{2} - 1}\) 2.

Multiply the numerator and denominator by \(\sqrt{2} + 1 \) 

\(\frac{\sqrt{2} + 1}{\sqrt{2} - 1} = \frac{(\sqrt{2} + 1)^2}{(\sqrt{2})^2 - 1^2} = \frac{2 + 2\sqrt{2} + 1}{2 - 1} = \frac{3 + 2\sqrt{2}}{1} = 3 + 2\sqrt{2}\)

So, the inequality becomes \(0 < x < 3 + 2\sqrt{2}\)

The set \( A \) consists of rational numbers \( x \in \mathbb{Q} \) that satisfy this inequality. Therefore,

The supremum \(\sup A \) of this set is \( 3 + 2\sqrt{2} \) , which is the least upper bound for \(x \) .

The infimum \(\inf A \) is 0 , since \(x \) approaches but never reaches 0 from the positive side.

The correct statement is \(\sup A = 3 + 2\sqrt{2} \)

Hence option 2) is correct.