Please include reasons for each step. **1.6.2 (1)***[Used in Example 1.6.3] Let T be the set defined in Equation 1.6.1 (1). Prove that T is a Dedekind cut.* **Example 1.6.3** To find a Dedekind cut that is not of the form given in Lemma 1.6.2, the simplest idea is to look at the set of all rational numbers greater than a real number that is not rational; the problem is how to describe such a set without making use of real numbers. In the case of the number \( x = \sqrt{2} \), there turns out to be a simple solution to this problem, as we will now see. We note first, however, that we have not yet formally defined what \( \sqrt{2} \) means, nor proved that there is such a real number, though we will do so in Theorem 2.6.9 and Definition 2.6.10. We have not yet proved that \( \sqrt{2} \) is not rational, a fact with which the reader is, at least informally, familiar; we will see a proof of this fact in Theorem 2.6.11. More precisely, it will be seen in that example that there is no rational number \( x \) such that \( x^2 = 2 \), and this last statement makes use only of rational numbers, so it is suited to our purpose at present. Nothing in our subsequent treatment of \( \sqrt{2} \) in Section 2.6 makes use of the current example, so it will not be circular reasoning for us to make use of these subsequently proved facts here.Let\[ T = \{ x \in \mathbb{Q} \mid x > 0 \text{ and } x^2 < 2 \}. \](1.6.1)It is seen by Exercise 1.6.2 (1) that \( T \) is a Dedekind cut, and by Part (2) of that exercise it is seen that if \( T \) has the form \( \{ x \in \mathbb{Q} \mid x > r \} \) for some \( r \in \mathbb{Q} \), then \( r^2 = 2 \). By Theorem 2.6.11 we know that there is no rational number \( x \) such that \( x^2 = 2 \), and this shows that \( T \) is a Dedekind cut that is not of the

let T be the set defined as T=x∈ℚ| x>0 and x2>2 claim- to prove that T is dedekind cut. it is known that a Dedekind cut is defined as a partition of rational numbers into two sets A and B such that all elements of A are less than all elements of B and the set A has no greatest element. now, defined a set B=x∈ℚ| x≤0 and x2<2 (i) we can see both the sets T and B are non empty. (ii) take compliment of set T Tc=x∈ℚ| x≤0 and x2<2 implies, B=Tc also intersection of both the sets B∩T=ϕ and union of both the sets B∪T=ϕ (iii) suppose that x,y∈ℚ be such that x<y and y∈B using definition of B=x∈ℚ| x≤0 and x2<2 since y∈B implies, y≤0 or y2<2 since x<y using above condition x≤0 or x2<2 implies, x∈B(iv) suppose x∈B implies, x≤0 or x2<2 if x≤0 then we can choose y=1 such that x<y and if x2<2 then we get x<2 now, by using density property of rationals we get there exist a rational y such that x<y<2 implies y2<2 hence we get for all element x∈B there exist y∈B such that x<y implies, set B has no largest element. hence proved set T is a Dedekind cut