A. प्रतिवर्ती और संक्रामी लेकिन सममित नहीं/Reflexive and transitive but not symmetric
Step 1
Concept
For every \(a\in A\), \(a\mid a\), and if \(a\mid b\), \(b\mid c\), then \(a\mid c\). But \(2\mid4\) while \(4\nmid2\), so it is not symmetric.
Step 2
Why this answer is correct
The correct answer is A. प्रतिवर्ती और संक्रामी लेकिन सममित नहीं / Reflexive and transitive but not symmetric. For every \(a\in A\), \(a\mid a\), and if \(a\mid b\), \(b\mid c\), then \(a\mid c\). But \(2\mid4\) while \(4\nmid2\), so it is not symmetric.
Step 3
Exam Tip
हर \(a\in A\) के लिए \(a\mid a\) और यदि \(a\mid b\), \(b\mid c\), तो \(a\mid c\)। पर \(2\mid4\) है लेकिन \(4\nmid2\), इसलिए सममित नहीं।
(a+b) is even exactly when (a) and (b) have the same parity. Hence the equivalence classes are the odd class ({1,3}) and the even class ({2,4}).
Step 2
Why this answer is correct
The correct answer is A. ({1,3}) और ({2,4}) / ({1,3}) and ({2,4}). (a+b) is even exactly when (a) and (b) have the same parity. Hence the equivalence classes are the odd class ({1,3}) and the even class ({2,4}).
Step 3
Exam Tip
(a+b) सम तभी होता है जब (a) और (b) दोनों समान parity के हों। इसलिए equivalence classes विषम ({1,3}) और सम ({2,4}) हैं।
(a+a) is even, (a+b) even implies (b+a) even, and same parity gives transitivity. In exams, identify the even and odd equivalence classes.
Step 2
Why this answer is correct
The correct answer is A. तुल्यता संबंध / Equivalence relation. (a+a) is even, (a+b) even implies (b+a) even, and same parity gives transitivity. In exams, identify the even and odd equivalence classes.
Step 3
Exam Tip
(a+a) सम है, (a+b) सम होने पर (b+a) सम है, और समान parity से संक्रामकता मिलती है। परीक्षा में इसे विषम और सम वर्गों से पहचानें।
\(A\times A\) has \(n^2\) ordered pairs, and a relation can be any subset of it. Hence the number of relations is \(2^{n^2}\).
Step 2
Why this answer is correct
The correct answer is A. \(2^{n^2}\). \(A\times A\) has \(n^2\) ordered pairs, and a relation can be any subset of it. Hence the number of relations is \(2^{n^2}\).
Step 3
Exam Tip
\(A\times A\) में \(n^2\) ordered pairs होते हैं और relation उसका कोई भी उपसमुच्चय हो सकता है। इसलिए कुल \(2^{n^2}\) संबंध बनते हैं।
\(A\times B\) has (mn) ordered pairs. Each pair may be included or excluded, so there are \(2^{mn}\) possible relations.
Step 2
Why this answer is correct
The correct answer is A. \(2^{mn}\). \(A\times B\) has (mn) ordered pairs. Each pair may be included or excluded, so there are \(2^{mn}\) possible relations.
Step 3
Exam Tip
\(A\times B\) में (mn) ordered pairs होते हैं। प्रत्येक pair relation में हो या न हो, इसलिए कुल \(2^{mn}\) विकल्प हैं।
A reflexive relation must contain ((1,1),(2,2),(3,3)). The remaining (9-3=6) pairs are optional, so the answer is \(2^6\).
Step 2
Why this answer is correct
The correct answer is A. \(2^6\). A reflexive relation must contain ((1,1),(2,2),(3,3)). The remaining (9-3=6) pairs are optional, so the answer is \(2^6\).
Step 3
Exam Tip
प्रतिवर्ती संबंध में ((1,1),(2,2),(3,3)) अनिवार्य हैं। बाकी (9-3=6) pairs वैकल्पिक हैं, इसलिए उत्तर \(2^6\) है।
There are \(4^2=16\) ordered pairs, and (4) diagonal pairs are compulsory. Thus (12) pairs are optional, giving \(2^{12}\).
Step 2
Why this answer is correct
The correct answer is A. \(2^{12}\). There are \(4^2=16\) ordered pairs, and (4) diagonal pairs are compulsory. Thus (12) pairs are optional, giving \(2^{12}\).
Step 3
Exam Tip
कुल ordered pairs \(4^2=16\) हैं और (4) diagonal pairs अनिवार्य हैं। अतः वैकल्पिक pairs (12) हैं, इसलिए \(2^{12}\)।
For a symmetric relation, (3) diagonal pairs and (3) unordered off-diagonal pair blocks are independently chosen. Thus there are (6) choices, giving \(2^6\).
Step 2
Why this answer is correct
The correct answer is A. \(2^6\). For a symmetric relation, (3) diagonal pairs and (3) unordered off-diagonal pair blocks are independently chosen. Thus there are (6) choices, giving \(2^6\).
Step 3
Exam Tip
सममित संबंध में (3) diagonal pairs स्वतंत्र हैं और (3) unordered off-diagonal pair blocks स्वतंत्र हैं। कुल स्वतंत्र चुनाव (6) हैं, इसलिए \(2^6\)।
For a symmetric relation, the number of independent choices is (\frac{4(4+1)}{2}=10). Hence the number of relations is \(2^{10}\).
Step 2
Why this answer is correct
The correct answer is A. \(2^{10}\). For a symmetric relation, the number of independent choices is (\frac{4(4+1)}{2}=10). Hence the number of relations is \(2^{10}\).
Step 3
Exam Tip
सममित संबंध में स्वतंत्र चुनावों की संख्या (\frac{4(4+1)}{2}=10) होती है। इसलिए कुल संबंध \(2^{10}\) हैं।
Reflexivity fixes the (5) diagonal pairs. Symmetry leaves only (\frac{5(5-1)}{2}=10) off-diagonal blocks free, so the answer is \(2^{10}\).
Step 2
Why this answer is correct
The correct answer is A. \(2^{10}\). Reflexivity fixes the (5) diagonal pairs. Symmetry leaves only (\frac{5(5-1)}{2}=10) off-diagonal blocks free, so the answer is \(2^{10}\).
Step 3
Exam Tip
प्रतिवर्ती होने से (5) diagonal pairs fixed हैं। सममिति में केवल (\frac{5(5-1)}{2}=10) off-diagonal blocks स्वतंत्र हैं, इसलिए \(2^{10}\)।
समुच्चय \(A=\{1,2,3\}\) पर relation \(R=\{(1,1),(2,2),(3,3),(1,2),(2,1),(2,3)\}\) है। (R) को सममित बनाने के लिए न्यूनतम कौन सा ordered pair जोड़ना होगा?
A. (R) संक्रामी है पर प्रतिवर्ती नहीं/(R) is transitive but not reflexive
Step 1
Concept
From ((1,2)) and ((2,3)), ((1,3)) is present, so the needed transitive condition holds. But ((1,1),(2,2),(3,3)) are absent, so it is not reflexive.
Step 2
Why this answer is correct
The correct answer is A. (R) संक्रामी है पर प्रतिवर्ती नहीं / (R) is transitive but not reflexive. From ((1,2)) and ((2,3)), ((1,3)) is present, so the needed transitive condition holds. But ((1,1),(2,2),(3,3)) are absent, so it is not reflexive.
Step 3
Exam Tip
((1,2)) और ((2,3)) से ((1,3)) मौजूद है, इसलिए जरूरी संक्रामक शर्त पूरी है। पर ((1,1),(2,2),(3,3)) नहीं हैं, इसलिए प्रतिवर्ती नहीं।
A. सममित लेकिन संक्रामी नहीं/Symmetric but not transitive
Step 1
Concept
Every present pair has its reverse, so the relation is symmetric. But ((1,2)) and ((2,3)) are present while ((1,3)) is not, so it is not transitive.
Step 2
Why this answer is correct
The correct answer is A. सममित लेकिन संक्रामी नहीं / Symmetric but not transitive. Every present pair has its reverse, so the relation is symmetric. But ((1,2)) and ((2,3)) are present while ((1,3)) is not, so it is not transitive.
Step 3
Exam Tip
हर मौजूद pair का reverse मौजूद है, इसलिए सममित है। पर ((1,2)) और ((2,3)) हैं लेकिन ((1,3)) नहीं है, इसलिए संक्रामी नहीं।
Since \(a-a=0\in\mathbb{Z}\), \(a-b\in\mathbb{Z}\) implies \(b-a\in\mathbb{Z}\), and the sum of integers is an integer. Hence it is an equivalence relation.
Step 2
Why this answer is correct
The correct answer is A. तुल्यता संबंध / Equivalence relation. Since \(a-a=0\in\mathbb{Z}\), \(a-b\in\mathbb{Z}\) implies \(b-a\in\mathbb{Z}\), and the sum of integers is an integer. Hence it is an equivalence relation.
Step 3
Exam Tip
\(a-a=0\in\mathbb{Z}\), \(a-b\in\mathbb{Z}\) से \(b-a\in\mathbb{Z}\), और पूर्णांकों का योग पूर्णांक है। इसलिए यह तुल्यता संबंध है।
Because \(7\equiv 2 \pmod{5}\), all integers with the same remainder are in its class. Always form an equivalence class from the relation condition.
Step 2
Why this answer is correct
The correct answer is A. \({x\in\mathbb{Z}:x\equiv 2 \pmod{5}}\). Because \(7\equiv 2 \pmod{5}\), all integers with the same remainder are in its class. Always form an equivalence class from the relation condition.
Step 3
Exam Tip
क्योंकि \(7\equiv 2 \pmod{5}\), इसलिए उसी शेषफल वाले सभी पूर्णांक class में होंगे। Equivalence class हमेशा relation की condition से बनाइए।
By same parity, the odd class is ({1,3}) and the even class is ({2,4}). A partition is the collection of equivalence classes.
Step 2
Why this answer is correct
The correct answer is A. ({{1,3},{2,4}}). By same parity, the odd class is ({1,3}) and the even class is ({2,4}). A partition is the collection of equivalence classes.
Step 3
Exam Tip
Same parity के अनुसार विषम ({1,3}) और सम ({2,4}) classes मिलती हैं। Partition हमेशा equivalence classes का समूह होता है।
A. यह तुल्यता संबंध है/It is an equivalence relation
Step 1
Concept
Since \(a^2=a^2\), equality is symmetric, and \(a^2=b^2\), \(b^2=c^2\) imply \(a^2=c^2\). Hence (R) is an equivalence relation.
Step 2
Why this answer is correct
The correct answer is A. यह तुल्यता संबंध है / It is an equivalence relation. Since \(a^2=a^2\), equality is symmetric, and \(a^2=b^2\), \(b^2=c^2\) imply \(a^2=c^2\). Hence (R) is an equivalence relation.
Step 3
Exam Tip
\(a^2=a^2\), equality symmetric होती है, और \(a^2=b^2\), \(b^2=c^2\) से \(a^2=c^2\)। इसलिए (R) तुल्यता संबंध है।
A. सममित लेकिन प्रतिवर्ती नहीं/Symmetric but not reflexive
Step 1
Concept
Since (\gcd(a,b)=\gcd(b,a)), the relation is symmetric. But (\gcd(2,2)=2\neq1), so it is not reflexive.
Step 2
Why this answer is correct
The correct answer is A. सममित लेकिन प्रतिवर्ती नहीं / Symmetric but not reflexive. Since (\gcd(a,b)=\gcd(b,a)), the relation is symmetric. But (\gcd(2,2)=2\neq1), so it is not reflexive.
Step 3
Exam Tip
(\gcd(a,b)=\gcd(b,a)), इसलिए relation सममित है। पर (\gcd(2,2)=2\neq1), इसलिए प्रतिवर्ती नहीं।
Because (a+b=b+a), reversing the pair keeps the same sum. But not every (a+a) is prime, so reflexivity is not guaranteed.
Step 2
Why this answer is correct
The correct answer is A. सममित / Symmetric. Because (a+b=b+a), reversing the pair keeps the same sum. But not every (a+a) is prime, so reflexivity is not guaranteed.
Step 3
Exam Tip
क्योंकि (a+b=b+a), pair उलटने पर भी योग वही रहता है। पर हर (a+a) prime नहीं होता, इसलिए reflexive निश्चित नहीं है।
A. सममित लेकिन प्रतिवर्ती नहीं/Symmetric but not reflexive
Step 1
Concept
If (a+b=5), then (b+a=5), so it is symmetric. But ((1,1)) is not in the relation because \(1+1\neq5\).
Step 2
Why this answer is correct
The correct answer is A. सममित लेकिन प्रतिवर्ती नहीं / Symmetric but not reflexive. If (a+b=5), then (b+a=5), so it is symmetric. But ((1,1)) is not in the relation because \(1+1\neq5\).
Step 3
Exam Tip
यदि (a+b=5), तो (b+a=5), इसलिए सममित है। पर ((1,1)) relation में नहीं क्योंकि \(1+1\neq5\)।
A. प्रतिवर्ती और संक्रामी लेकिन सममित नहीं/Reflexive and transitive but not symmetric
Step 1
Concept
Every \(a\leq a\) is true, and \(a\leq b\leq c\) implies \(a\leq c\). But \(2\leq3\) does not imply \(3\leq2\).
Step 2
Why this answer is correct
The correct answer is A. प्रतिवर्ती और संक्रामी लेकिन सममित नहीं / Reflexive and transitive but not symmetric. Every \(a\leq a\) is true, and \(a\leq b\leq c\) implies \(a\leq c\). But \(2\leq3\) does not imply \(3\leq2\).
Step 3
Exam Tip
हर \(a\leq a\) सत्य है और \(a\leq b\leq c\) से \(a\leq c\)। लेकिन \(2\leq3\) के बाद \(3\leq2\) सत्य नहीं।
Every \(A\subseteq A\), and the subset relation is antisymmetric and transitive. Do not treat it like equality because it is generally not symmetric.
Step 2
Why this answer is correct
The correct answer is A. आंशिक क्रम संबंध / Partial order relation. Every \(A\subseteq A\), and the subset relation is antisymmetric and transitive. Do not treat it like equality because it is generally not symmetric.
Step 3
Exam Tip
हर \(A\subseteq A\), और subset relation antisymmetric तथा transitive होता है। इसे equality जैसा नहीं मानें क्योंकि यह generally symmetric नहीं होता।
\((1,2)\in R\) but \((2,1)\notin R\), so it is not symmetric. Since all diagonal pairs are present, it is reflexive.
Step 2
Why this answer is correct
The correct answer is A. सममित / Symmetric. \((1,2)\in R\) but \((2,1)\notin R\), so it is not symmetric. Since all diagonal pairs are present, it is reflexive.
Step 3
Exam Tip
\((1,2)\in R\) है पर \((2,1)\notin R\), इसलिए सममित नहीं है। Diagonal pairs होने से reflexive है।
If ((a,b)) with \(a\neq b\) occurs, symmetry forces ((b,a)), contradicting antisymmetry. Hence only diagonal pairs can occur.
Step 2
Why this answer is correct
The correct answer is A. कोई भी नहीं / None. If ((a,b)) with \(a\neq b\) occurs, symmetry forces ((b,a)), contradicting antisymmetry. Hence only diagonal pairs can occur.
Step 3
Exam Tip
यदि ((a,b)) और \(a\neq b\) हो, तो symmetry से ((b,a)) भी होगा, जो antisymmetry का विरोध करेगा। इसलिए केवल diagonal pairs संभव हैं।
In the order \( \leq \), the greatest element is one that every element is less than or equal to. Here \(a\leq4\) for every \(a\in A\), so (4) is greatest.
Step 2
Why this answer is correct
The correct answer is A. (4). In the order \( \leq \), the greatest element is one that every element is less than or equal to. Here \(a\leq4\) for every \(a\in A\), so (4) is greatest.
Step 3
Exam Tip
Order \( \leq \) में greatest element वह है जिससे सभी elements छोटे या बराबर हों। यहाँ हर \(a\in A\) के लिए \(a\leq4\), इसलिए (4) greatest है।
A least element must divide every element. Here neither \(2\mid3\) nor \(3\mid2\), so no least element exists.
Step 2
Why this answer is correct
The correct answer is A. कोई नहीं / None. A least element must divide every element. Here neither \(2\mid3\) nor \(3\mid2\), so no least element exists.
Step 3
Exam Tip
Least element को सभी elements को divide करना चाहिए। यहाँ न \(2\mid3\) और न \(3\mid2\), इसलिए कोई least element नहीं है।
A. (1) least और (8) greatest है/(1) is least and (8) is greatest
Step 1
Concept
(1) divides every element, and every element divides (8). In divisibility order, identify least and greatest by the direction of divisibility.
Step 2
Why this answer is correct
The correct answer is A. (1) least और (8) greatest है / (1) is least and (8) is greatest. (1) divides every element, and every element divides (8). In divisibility order, identify least and greatest by the direction of divisibility.
Step 3
Exam Tip
(1) सभी को divide करता है और सभी (8) को divide करते हैं। Divisibility order में least और greatest को direction से पहचानें।
A. यह प्रतिवर्ती, सममित और संक्रामी है/It is reflexive, symmetric, and transitive
Step 1
Concept
In the universal relation, every possible ordered pair is present. Hence reflexive, symmetric, and transitive properties all hold automatically.
Step 2
Why this answer is correct
The correct answer is A. यह प्रतिवर्ती, सममित और संक्रामी है / It is reflexive, symmetric, and transitive. In the universal relation, every possible ordered pair is present. Hence reflexive, symmetric, and transitive properties all hold automatically.
Step 3
Exam Tip
Universal relation में हर possible ordered pair मौजूद होता है। इसलिए reflexive, symmetric और transitive तीनों properties स्वतः पूरी होती हैं।
A. यह सममित और संक्रामी है पर प्रतिवर्ती नहीं/It is symmetric and transitive but not reflexive
Step 1
Concept
The empty relation has no counterexample, so symmetry and transitivity are vacuously true. But \((1,1)\notin R\), so it is not reflexive.
Step 2
Why this answer is correct
The correct answer is A. यह सममित और संक्रामी है पर प्रतिवर्ती नहीं / It is symmetric and transitive but not reflexive. The empty relation has no counterexample, so symmetry and transitivity are vacuously true. But \((1,1)\notin R\), so it is not reflexive.
Step 3
Exam Tip
Empty relation में कोई counterexample नहीं, इसलिए symmetry और transitivity vacuously true हैं। पर \((1,1)\notin R\), इसलिए reflexive नहीं।
A. तुल्यता संबंध और आंशिक क्रम दोनों/Both equivalence relation and partial order
Step 1
Concept
The identity relation is reflexive, symmetric, and transitive, so it is an equivalence relation. It is also reflexive, antisymmetric, and transitive, so it is a partial order.
Step 2
Why this answer is correct
The correct answer is A. तुल्यता संबंध और आंशिक क्रम दोनों / Both equivalence relation and partial order. The identity relation is reflexive, symmetric, and transitive, so it is an equivalence relation. It is also reflexive, antisymmetric, and transitive, so it is a partial order.
Step 3
Exam Tip
Identity relation reflexive, symmetric और transitive है, इसलिए equivalence relation है। यह reflexive, antisymmetric और transitive भी है, इसलिए partial order भी है।
A. \(R^{-1}\) भी reflexive है/\(R^{-1}\) is also reflexive
Step 1
Concept
A reflexive (R) contains every ((a,a)), and the inverse of ((a,a)) is again ((a,a)). Therefore \(R^{-1}\) remains reflexive.
Step 2
Why this answer is correct
The correct answer is A. \(R^{-1}\) भी reflexive है / \(R^{-1}\) is also reflexive. A reflexive (R) contains every ((a,a)), and the inverse of ((a,a)) is again ((a,a)). Therefore \(R^{-1}\) remains reflexive.
Step 3
Exam Tip
Reflexive (R) में हर ((a,a)) होता है, और उसका inverse भी ((a,a)) ही है। इसलिए \(R^{-1}\) reflexive रहता है।
In a symmetric relation, \((a,b)\in R\) implies \((b,a)\in R\), so the inverse has the same pairs. Hence \(R=R^{-1}\).
Step 2
Why this answer is correct
The correct answer is A. \(R=R^{-1}\). In a symmetric relation, \((a,b)\in R\) implies \((b,a)\in R\), so the inverse has the same pairs. Hence \(R=R^{-1}\).
Step 3
Exam Tip
Symmetric relation में \((a,b)\in R\) से \((b,a)\in R\), इसलिए inverse में वही pairs मिलते हैं। अतः \(R=R^{-1}\)।
The classes are ({1,4},{2},{3}), so the number of pairs is \(2^2+1^2+1^2=6\). Add the squares of the sizes of the equivalence classes.
Step 2
Why this answer is correct
The correct answer is A. (6). The classes are ({1,4},{2},{3}), so the number of pairs is \(2^2+1^2+1^2=6\). Add the squares of the sizes of the equivalence classes.
Step 3
Exam Tip
Classes ({1,4},{2},{3}) हैं, इसलिए pairs की संख्या \(2^2+1^2+1^2=6\) है। Equivalence classes के sizes के squares जोड़ें।
A. प्रतिवर्ती और सममित लेकिन संक्रामी नहीं/Reflexive and symmetric but not transitive
Step 1
Concept
Since \(|a-a|=0\leq1\) and (|a-b|=|b-a|), it is reflexive and symmetric. But (1R2) and (2R3) hold while (1R3) does not.
Step 2
Why this answer is correct
The correct answer is A. प्रतिवर्ती और सममित लेकिन संक्रामी नहीं / Reflexive and symmetric but not transitive. Since \(|a-a|=0\leq1\) and (|a-b|=|b-a|), it is reflexive and symmetric. But (1R2) and (2R3) hold while (1R3) does not.
Step 3
Exam Tip
\(|a-a|=0\leq1\) और (|a-b|=|b-a|), इसलिए reflexive और symmetric है। लेकिन (1R2) और (2R3) हैं पर (1R3) नहीं।
A. अप्रतिवर्ती और संक्रामी/Irreflexive and transitive
Step 1
Concept
Since (a-a=0), (aRa) is never true, so it is irreflexive. If (a>b) and (b>c), then (a>c), so it is transitive.
Step 2
Why this answer is correct
The correct answer is A. अप्रतिवर्ती और संक्रामी / Irreflexive and transitive. Since (a-a=0), (aRa) is never true, so it is irreflexive. If (a>b) and (b>c), then (a>c), so it is transitive.
Step 3
Exam Tip
(a-a=0) होने से (aRa) कभी सत्य नहीं, इसलिए irreflexive है। यदि (a>b) और (b>c), तो (a>c), इसलिए transitive है।
The function \(x^3\) is strictly increasing, so \(a^3\leq b^3\) is exactly equivalent to \(a\leq b\). Use monotonicity to identify relation properties quickly.
Step 2
Why this answer is correct
The correct answer is A. \(a\leq b\). The function \(x^3\) is strictly increasing, so \(a^3\leq b^3\) is exactly equivalent to \(a\leq b\). Use monotonicity to identify relation properties quickly.
Step 3
Exam Tip
Function \(x^3\) strictly increasing है, इसलिए \(a^3\leq b^3\) exactly \(a\leq b\) के बराबर है। Monotonicity से relation की property जल्दी पहचानें।
A. सममित लेकिन प्रतिवर्ती नहीं/Symmetric but not reflexive
Step 1
Concept
Addition is commutative, so the relation is symmetric. But (a+a=2a) is always even, so (aRa) never holds.
Step 2
Why this answer is correct
The correct answer is A. सममित लेकिन प्रतिवर्ती नहीं / Symmetric but not reflexive. Addition is commutative, so the relation is symmetric. But (a+a=2a) is always even, so (aRa) never holds.
Step 3
Exam Tip
योग commutative है, इसलिए relation symmetric है। पर (a+a=2a) हमेशा सम है, इसलिए (aRa) कभी नहीं होता।
A. सममित लेकिन संक्रामी नहीं/Symmetric but not transitive
Step 1
Concept
If \(a\neq b\), then \(b\neq a\), so it is symmetric. But (1R2) and (2R1) hold while (1R1) does not, so it is not transitive.
Step 2
Why this answer is correct
The correct answer is A. सममित लेकिन संक्रामी नहीं / Symmetric but not transitive. If \(a\neq b\), then \(b\neq a\), so it is symmetric. But (1R2) and (2R1) hold while (1R1) does not, so it is not transitive.
Step 3
Exam Tip
यदि \(a\neq b\), तो \(b\neq a\), इसलिए symmetric है। लेकिन (1R2) और (2R1) हैं, फिर भी (1R1) नहीं, इसलिए transitive नहीं।
The given pairs are exactly those where the first element is less than or equal to the second. Thus it is the usual order \(\leq\) on (A).
Step 2
Why this answer is correct
The correct answer is A. \(a\leq b\). The given pairs are exactly those where the first element is less than or equal to the second. Thus it is the usual order \(\leq\) on (A).
Step 3
Exam Tip
दिए गए pairs वही हैं जिनमें पहला अवयव दूसरे से छोटा या बराबर है। इसलिए यह (A) पर usual order \(\leq\) है।
In a reflexive closure, ((a,a)) must be present for every \(a\in A\). Therefore all three diagonal pairs are added.
Step 2
Why this answer is correct
The correct answer is A. ((1,1),(2,2),(3,3)). In a reflexive closure, ((a,a)) must be present for every \(a\in A\). Therefore all three diagonal pairs are added.
Step 3
Exam Tip
Reflexive closure में हर \(a\in A\) के लिए ((a,a)) होना जरूरी है। इसलिए तीनों diagonal pairs जोड़े जाएंगे।
In a symmetric closure, ((b,a)) is added for every ((a,b)). Thus ((2,1)) is needed for ((1,2)), and ((3,2)) for ((2,3)).
Step 2
Why this answer is correct
The correct answer is A. ((2,1),(3,2)). In a symmetric closure, ((b,a)) is added for every ((a,b)). Thus ((2,1)) is needed for ((1,2)), and ((3,2)) for ((2,3)).
Step 3
Exam Tip
Symmetric closure में हर ((a,b)) के साथ ((b,a)) जोड़ा जाता है। इसलिए ((1,2)) के लिए ((2,1)) और ((2,3)) के लिए ((3,2)) चाहिए।
A. हाँ, क्योंकि कोई distinct reverse pair साथ में नहीं है/Yes, because no distinct reverse pair appears together
Step 1
Concept
Antisymmetry is not affected by diagonal pairs. It fails only when ((a,b)) and ((b,a)) both occur for \(a\neq b\).
Step 2
Why this answer is correct
The correct answer is A. हाँ, क्योंकि कोई distinct reverse pair साथ में नहीं है / Yes, because no distinct reverse pair appears together. Antisymmetry is not affected by diagonal pairs. It fails only when ((a,b)) and ((b,a)) both occur for \(a\neq b\).
Step 3
Exam Tip
Antisymmetry को diagonal pairs से समस्या नहीं होती। समस्या तभी होती है जब \(a\neq b\) पर ((a,b)) और ((b,a)) दोनों हों।
Having the same greatest prime factor behaves like equality, so it is reflexive, symmetric, and transitive. Such same invariant conditions often give equivalence relations.
Step 2
Why this answer is correct
The correct answer is A. तुल्यता संबंध / Equivalence relation. Having the same greatest prime factor behaves like equality, so it is reflexive, symmetric, and transitive. Such same invariant conditions often give equivalence relations.
Step 3
Exam Tip
एक ही greatest prime factor होना equality जैसी condition है, इसलिए reflexive, symmetric और transitive है। ऐसी same invariant वाली conditions अक्सर equivalence relation देती हैं।
(xRa) means \(x-a\in\mathbb{Q}\), so (x=a+q) where \(q\in\mathbb{Q}\). When writing a class, solve the relation condition for the variable.
Step 2
Why this answer is correct
The correct answer is A. \([a]={a+q:q\in\mathbb{Q}}\). (xRa) means \(x-a\in\mathbb{Q}\), so (x=a+q) where \(q\in\mathbb{Q}\). When writing a class, solve the relation condition for the variable.
Step 3
Exam Tip
(xRa) का अर्थ \(x-a\in\mathbb{Q}\), यानी (x=a+q) जहाँ \(q\in\mathbb{Q}\)। Class लिखते समय variable को relation condition से निकालें।
A. सममित लेकिन प्रतिवर्ती नहीं/Symmetric but not reflexive
Step 1
Concept
Since (a+b=b+a), the relation is symmetric. But \((4,4)\notin R\) because \(4+4\leq6\) is false, so it is not reflexive.
Step 2
Why this answer is correct
The correct answer is A. सममित लेकिन प्रतिवर्ती नहीं / Symmetric but not reflexive. Since (a+b=b+a), the relation is symmetric. But \((4,4)\notin R\) because \(4+4\leq6\) is false, so it is not reflexive.
Step 3
Exam Tip
(a+b=b+a), इसलिए relation symmetric है। पर \((4,4)\notin R\) क्योंकि \(4+4\leq6\) गलत है, इसलिए reflexive नहीं।
A. \(R\cap S\) भी equivalence relation है/\(R\cap S\) is also an equivalence relation
Step 1
Concept
The intersection retains common diagonal pairs, and symmetry and transitivity are also preserved. Hence the intersection of equivalence relations is again an equivalence relation.
Step 2
Why this answer is correct
The correct answer is A. \(R\cap S\) भी equivalence relation है / \(R\cap S\) is also an equivalence relation. The intersection retains common diagonal pairs, and symmetry and transitivity are also preserved. Hence the intersection of equivalence relations is again an equivalence relation.
Step 3
Exam Tip
Intersection में common diagonal pairs रहते हैं और symmetry तथा transitivity भी preserve होती हैं। इसलिए equivalence relations का intersection फिर equivalence relation होता है।
