Consider the following statements regarding Logic Gates:
1. NAND and NOR gates are universal gates capable of implementing any Boolean function.
2. XOR gate follows associative law under all conditions of logic operations.
3. AND gate operation is commutative and associative in Boolean algebra.
Which of the above statements is/are correct?
Answer & step-by-step solution
Boolean algebra has operation-specific laws. AND and OR satisfy commutative and associative laws; XOR also satisfies commutative and associative laws, although it is not idempotent (A⊕A=0). Key relation: NAND universality: NOT A=A NAND A; AND from double NAND; OR via De Morgan. XOR associativity: A⊕B⊕C is independent of grouping.
Exam focus: Do not confuse XOR’s non-idempotent property with associativity. Common trap: Assuming XOR is non-associative because its output changes with parity.
Statement-wise Verification
- Statement 1 - Correct.
NAND and NOR gates are universal gates capable of implementing any Boolean function.
Any Boolean function can be constructed using NAND alone or NOR alone. - Statement 2 - Correct.
XOR gate follows associative law under all conditions of logic operations.
XOR is associative, so A⊕(B⊕C)=(A⊕B)⊕C for binary Boolean operands. - Statement 3 - Correct.
AND gate operation is commutative and associative in Boolean algebra.
Boolean algebra has operation-specific laws.
Core Concept
Boolean algebra has operation-specific laws. AND and OR satisfy commutative and associative laws; XOR also satisfies commutative and associative laws, although it is not idempotent (A⊕A=0). Universal-gate status means functional completeness, not that every gate has the same truth table.
Formula / Key Relationship
NAND universality: NOT A=A NAND A; AND from double NAND; OR via De Morgan. XOR associativity: A⊕B⊕C is independent of grouping.
Step-by-Step Check
Truth-table or algebraic verification confirms each statement. Therefore D (1,2,3) is the complete answer.
Why the Other Options Are Wrong
Option A is incorrect because it omits correct statement(s) 3. Option B is incorrect because it omits correct statement(s) 2. Option C is incorrect because it omits correct statement(s) 1.
Answer-Key Verification Note
The uploaded provisional key marks B . After independent technical verification, the defensible answer is D . This discrepancy is stated explicitly rather than silently copying the provisional key.
Exam Shortcut / Approach
Do not confuse XOR’s non-idempotent property with associativity.
Common Mistake
Assuming XOR is non-associative because its output changes with parity.
Quick Revision
Universal Gates and Algebraic Properties is linked with Boolean Operators, Logic Gates and Digital Electronics. Do not confuse XOR’s non-idempotent property with associativity.
लॉजिक गेट्स के संबंध में निम्नलिखित कथनों पर विचार कीजिए:
1. NAND और NOR गेट सार्वभौमिक (universal) गेट हैं जो कोई भी बूलियन फलन बना सकते हैं।
2. XOR गेट सभी परिस्थितियों में साहचर्य (associative) नियम का पालन करता है।
3. बूलियन बीजगणित में AND गेट क्रिया क्रमविनिमेय (commutative) और साहचर्य है।
उपरोक्त में से कौन-सा/से कथन सही है/हैं?
उत्तर और चरणबद्ध हल
AND और OR क्रमविनिमेय और साहचर्य नियम मानते हैं; XOR भी क्रमविनिमेय और साहचर्य है, यद्यपि यह इडेम्पोटेंट नहीं है (A⊕A=0)। मुख्य संबंध: NAND से NOT A=A NAND A; डी-मॉर्गन से OR; A⊕B⊕C समूहन से स्वतंत्र है।
परीक्षा फोकस: XOR के गैर-इडेम्पोटेंट गुण को साहचर्यता से न मिलाएँ।
विस्तृत हल (मूल भाषा में)
Statement-wise Verification
- Statement 1 - Correct.
NAND and NOR gates are universal gates capable of implementing any Boolean function.
Any Boolean function can be constructed using NAND alone or NOR alone. - Statement 2 - Correct.
XOR gate follows associative law under all conditions of logic operations.
XOR is associative, so A⊕(B⊕C)=(A⊕B)⊕C for binary Boolean operands. - Statement 3 - Correct.
AND gate operation is commutative and associative in Boolean algebra.
Boolean algebra has operation-specific laws.
Core Concept
Boolean algebra has operation-specific laws. AND and OR satisfy commutative and associative laws; XOR also satisfies commutative and associative laws, although it is not idempotent (A⊕A=0). Universal-gate status means functional completeness, not that every gate has the same truth table.
Formula / Key Relationship
NAND universality: NOT A=A NAND A; AND from double NAND; OR via De Morgan. XOR associativity: A⊕B⊕C is independent of grouping.
Step-by-Step Check
Truth-table or algebraic verification confirms each statement. Therefore D (1,2,3) is the complete answer.
Why the Other Options Are Wrong
Option A is incorrect because it omits correct statement(s) 3. Option B is incorrect because it omits correct statement(s) 2. Option C is incorrect because it omits correct statement(s) 1.
Answer-Key Verification Note
The uploaded provisional key marks B . After independent technical verification, the defensible answer is D . This discrepancy is stated explicitly rather than silently copying the provisional key.
Exam Shortcut / Approach
Do not confuse XOR’s non-idempotent property with associativity.
Common Mistake
Assuming XOR is non-associative because its output changes with parity.
Quick Revision
Universal Gates and Algebraic Properties is linked with Boolean Operators, Logic Gates and Digital Electronics. Do not confuse XOR’s non-idempotent property with associativity.
લોજિક ગેટ્સ સંબંધિત નીચેના વિધાનો ધ્યાનમાં લો:
1. NAND અને NOR ગેટ સાર્વત્રિક (universal) ગેટ છે જે કોઈ પણ બુલિયન વિધેય બનાવી શકે છે.
2. XOR ગેટ તમામ સ્થિતિમાં જૂથ (associative) નિયમનું પાલન કરે છે.
3. બુલિયન બીજગણિતમાં AND ગેટ ક્રિયા ક્રમનિરપેક્ષ (commutative) અને જૂથ નિયમ પાળે છે.
ઉપરોક્તમાંથી કયું/કયા વિધાન સાચું/સાચા છે?
જવાબ અને પગલાવાર ઉકેલ
AND અને OR ક્રમનિરપેક્ષ અને જૂથ નિયમ પાળે છે; XOR પણ ક્રમનિરપેક્ષ અને જૂથ નિયમ પાળે છે, જોકે તે ઇડેમ્પોટન્ટ નથી (A⊕A=0). મુખ્ય સંબંધ: NAND થી NOT A=A NAND A; ડી-મોર્ગનથી OR; A⊕B⊕C જૂથબંધીથી સ્વતંત્ર છે.
પરીક્ષા મુદ્દો: XOR ના બિન-ઇડેમ્પોટન્ટ ગુણને જૂથ નિયમ સાથે ગૂંચવશો નહીં.
વિગતવાર ઉકેલ (મૂળ ભાષામાં)
Statement-wise Verification
- Statement 1 - Correct.
NAND and NOR gates are universal gates capable of implementing any Boolean function.
Any Boolean function can be constructed using NAND alone or NOR alone. - Statement 2 - Correct.
XOR gate follows associative law under all conditions of logic operations.
XOR is associative, so A⊕(B⊕C)=(A⊕B)⊕C for binary Boolean operands. - Statement 3 - Correct.
AND gate operation is commutative and associative in Boolean algebra.
Boolean algebra has operation-specific laws.
Core Concept
Boolean algebra has operation-specific laws. AND and OR satisfy commutative and associative laws; XOR also satisfies commutative and associative laws, although it is not idempotent (A⊕A=0). Universal-gate status means functional completeness, not that every gate has the same truth table.
Formula / Key Relationship
NAND universality: NOT A=A NAND A; AND from double NAND; OR via De Morgan. XOR associativity: A⊕B⊕C is independent of grouping.
Step-by-Step Check
Truth-table or algebraic verification confirms each statement. Therefore D (1,2,3) is the complete answer.
Why the Other Options Are Wrong
Option A is incorrect because it omits correct statement(s) 3. Option B is incorrect because it omits correct statement(s) 2. Option C is incorrect because it omits correct statement(s) 1.
Answer-Key Verification Note
The uploaded provisional key marks B . After independent technical verification, the defensible answer is D . This discrepancy is stated explicitly rather than silently copying the provisional key.
Exam Shortcut / Approach
Do not confuse XOR’s non-idempotent property with associativity.
Common Mistake
Assuming XOR is non-associative because its output changes with parity.
Quick Revision
Universal Gates and Algebraic Properties is linked with Boolean Operators, Logic Gates and Digital Electronics. Do not confuse XOR’s non-idempotent property with associativity.
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