क्यों मायने रखता है

उत्तर प्रदेश के वित्तीय परिवर्तन, बजट वृद्धि की दिशा (वर्ष 2017 में 3 लाख करोड़ रुपये से बढ़कर वर्ष 2026 में 9 लाख करोड़ रुपये होना), और एक्सप्रेसवे पर केंद्रित पूंजीगत व्यय (capital expenditure) को समझना UPPSC Prelims के उन प्रश्नों के लिए अत्यधिक प्रासंगिक है, जो राज्य-स्तरीय बजट आंकड़ों, आर्थिक विकास और प्रमुख बुनियादी ढांचा पहलों को संबोधित करते हैं।of_the_following_options_are_incorrect? I. (e, g) ∈ R II. (g, e) ∈ R III. (g, g) ∈ R IV. (e, e) ∈ R Let's test these elements. Set S = {e, f, g, h}. e = [0, 1] f = [0, 0.5] g = [0.5, 1] h = [0.2, 0.7] Relation R is defined by: xRy ⇔ either (x ⊆ y) or (y ⊆ x). Let's check the options: I. (e, g) ∈ R Is e ⊆ g or g ⊆ e? e = [0, 1], g = [0.5, 1] Clearly, g ⊆ e since [0.5, 1] ⊆ [0, 1]. So, since g ⊆ e, (e, g) ∈ R (or is the relation symmetric? "either x ⊆ y or y ⊆ x" is symmetric). Let's check the definition of R: xRy iff x ⊆ y or y ⊆ x. For x=e, y=g: Does e ⊆ g? No, [0, 1] is not a subset of [0.5, 1]. Does g ⊆ e? Yes, [0.5, 1] is a subset of [0, 1]. Since one of the conditions is met (g ⊆ e), we have eRg. Thus, (e, g) ∈ R is true. Wait, let's check II: (g, e) ∈ R. By symmetry, if (e, g) ∈ R, then (g, e) ∈ R. Indeed, for x=g, y=e: Does g ⊆ e? Yes. Thus, (g, e) ∈ R is true. III. (g, g) ∈ R For x=g, y=g: Does g ⊆ g? Yes, every set is a subset of itself. Thus, (g, g) ∈ R is true. IV. (e, e) ∈ R For x=e, y=e: Does e ⊆ e? Yes. Thus, (e, e) ∈ R is true. Wait, are all of them correct? Let me re-read. Is there any typo in my understanding? Ah! "Which of the following options are incorrect?" But all four options I evaluated are correct according to the definition. Wait! Let me re-read the definitions of the sets. S = {e, f, g, h} where e = [0, 1], f = [0, 0.5], g = [0.5, 1], h = [0.2, 0.7]. Let me re-check. Is there any option that is NOT in R? Wait, if all are correct, maybe I made a mistake. Let me check the relation definition again. "xRy iff either x ⊆ y or y ⊆ x" Let's check if there is another interpretation. Could the question be about reflexivity? R is reflexive, so (x, x) is always in R. Thus III and IV are definitely correct. What about I and II? e = [0, 1] g = [0.5, 1] g ⊆ e is true because [0.5, 1] is a subinterval of [0, 1]. So both (e, g) and (g, e) must be in R. Why would any of them be incorrect? Wait! Let me search for the exact question from the exam. "Let S = {e, f, g, h} be a set of four intervals on the real line: e = [0, 1], f = [0, 0.5], g = [0.5, 1], h = [0.2, 0.7]... xRy if x ⊆ y or y ⊆ x." Actually, some options in the original GATE question might have been different! Ah! In the actual GATE question: The options were: (A) R is reflexive and transitive but not symmetric (B) R is reflexive and symmetric but not transitive (C) R is symmetric and transitive but not reflexive (D) R is an equivalence relation Oh! The prompt says: "Which of the following options are incorrect? I. (e, g) \in R II. (g, e) \in R III. (g, g) \in R IV. (e, e) \in R" Wait, if this is a modified version of the question, or if there's a typo in the options provided in the prompt? Wait! Let me check the definition of the intervals. e = [0, 1], f = [0, 0.5], g = [0.5, 1], h = [0.2, 0.7] Wait, is g = [0.5, 1] or maybe [0.6, 1]? In either case, g ⊆ e is true. What if g is NOT a subset of e? No, 0.5 >= 0 and 1 <= 1, so [0.5, 1] is definitely a subset of [0, 1]. What if the question meant "Which of the following options is/are correct?" and there's a "not" in one of them in the original test this prompt was taken from? Wait, could it be that R is defined as "x ⊂ y" (strict subset)? If xRy iff x ⊂ y or y ⊂ x: Then (g, g) and (e, e) would NOT be in R, because g is not a strict subset of g. But the standard symbol "⊆" means subset (which can be equal). If "⊆" is used, then: (e, e) and (g, g) are in R. And since g ⊆ e, (e, g) and (g, e) are in R. Thus all four are correct. Wait, let me think if there is any other element, say h and f. h = [0.2, 0.7], f = [0, 0.5]. Neither is a subset of the other, so (h, f) ∉ R. But the options only have e and g! e and g are [0, 1] and [0.5, 1]. Clearly [0.5, 1] ⊆ [0, 1]. So g ⊆ e. Thus (e, g) and (g, e) are in R. And since ⊆ is reflexive, (e, e) and (g, g) are in R. Therefore, NONE of the options I, II, III, IV are incorrect! Wait, could there be a typo in the prompt's options? Maybe "none of the above" or "none"? Since the question asks "Which of the following options are incorrect?", and all of them are actually correct (i.e., they belong to R), the answer is "None of the given options are incorrect" or "None". Let me double check if there's any other way to read "xRy iff x ⊆ y or y ⊆ x". If R is defined as above: - Reflexivity: For any x ∈ S, x ⊆ x is true, so xRx is true. Thus (e, e) ∈ R and (g, g) ∈ R are correct. - Since g = [0.5, 1] and e = [0, 1], we have g ⊆ e. - Therefore, by definition of R: - eRe holds because e ⊆ e. - gRg holds because g ⊆ g. - eRg holds because g ⊆ e. - gRe holds because g ⊆ e. So all four statements: I. (e, g) ∈ R II. (g, e) ∈ R III. (g, g) ∈ R IV. (e, e) ∈ R are mathematically TRUE. Thus, none of the options are incorrect. I will state this clearly, showing the step-by-step verification of each of the four statements to prove that they are all correct (i.e., they all belong to R), and therefore there are no incorrect options among the four. To find which of the given options are incorrect, we must test each statement against the definition of the relation $R$. ### **Given:** * **Set of intervals:** $S = \{e, f, g, h\}$ * $e = [0, 1]$ * $f = [0, 0.5]$ * $g = [0.5, 1]$ * $h = [0.2, 0.7]$ * **Relation $R$:** $$xRy \iff (x \subseteq y) \text{ or } (y \subseteq x)$$ --- ### **Testing the Options:** 1. **For Statement I: $(e, g) \in R$** * By definition, $(e, g) \in R \iff (e \subseteq g)$ or $(g \subseteq e)$. * Since $e = [0, 1]$ and $g = [0.5, 1]$, every element in $g$ is also in $e$, which means $g \subseteq e$ is **True**. * Therefore, $(e, g) \in R$. This statement is **correct**. 2. **For Statement II: $(g, e) \in R$** * By definition, $(g, e) \in R \iff (g \subseteq e)$ or $(e \subseteq g)$. * As established above, $g \subseteq e$ is **True**. * Therefore, $(g, e) \in R$. This statement is **correct**. 3. **For Statement III: $(g, g) \in R$** * By definition, $(g, g) \in R \iff (g \subseteq g)$. * Any set is a subset of itself, so $g \subseteq g$ is **True**. * Therefore, $(g, g) \in R$. This statement is **correct**. 4. **For Statement IV: $(e, e) \in R$** * By definition, $(e, e) \in R \iff (e \subseteq e)$. * Since $e \subseteq e$ is always **True**, $(e, e) \in R$. This statement is **correct**. --- ### **Conclusion:** All four statements (I, II, III, and IV) are **correct**. Therefore, **none of the options are incorrect**.