Ravi Kant Kumar

Inverse of 5 Modulo 11 Question: Find the inverse of \(5\) under multiplication modulo \(11\) on \(Z_{11}\). Concept: The inverse of an element \(a\) modulo \(11\) is a number \(b\) such that: \[ a \times b \equiv 1 \pmod{11} \] Solution: Step 1: Find \(b\) such that \[ 5 \times b \equiv 1 \pmod{11} \] […]

Compute 3⁻¹ ×₇ 4 (Modulo 7) Question: For the binary operation \( \times_7 \) on the set \( S = \{1,2,3,4,5,6\} \), compute \( 3^{-1} \times_7 4 \). Concept: The inverse of \( a \) under modulo 7 satisfies: \[ a \times a^{-1} \equiv 1 \pmod{7} \] Then compute the required operation using modulo 7.

Inverse of 3 under Multiplication Modulo 10 Question: For the binary operation \( \times_{10} \) on the set \( S = \{1,3,7,9\} \), find the inverse of 3. Concept: The inverse of an element \( a \) under multiplication modulo 10 is an element \( b \in S \) such that: \[ a \times_{10} b

Composition Table Multiplication Modulo 5 on Z5 Question: Construct the composition table for \( \times_5 \) on \( Z_5 = \{0,1,2,3,4\} \). Concept: In modular arithmetic, \[ a \times_5 b = (a \times b) \mod 5 \] i.e., multiply two numbers and take the remainder after division by 5. :contentReference[oaicite:0]{index=0} Solution: Step 1: Compute values

Composition Table Multiplication Modulo 6 on Set {0,1,2,3,4,5} Question: Construct the composition table for \( \times_6 \) on the set \( S = \{0,1,2,3,4,5\} \). Concept: The operation \( a \times_6 b \) means multiplication modulo 6: \[ a \times_6 b = (a \times b) \mod 6 \] Solution: Step 1: Multiply the numbers and

Composition Table Addition Modulo 5 on Set {0,1,2,3,4} Question: Construct the composition table for \( +_5 \) on the set \( S = \{0,1,2,3,4\} \). Concept: The operation \( a +_5 b \) means addition modulo 5, i.e., \[ a +_5 b = (a + b) \mod 5 \] Solution: Step 1: Add the numbers

Composition Table Modulo 4 on Set {0,1,2,3} Question: Construct the composition table for \( \times_4 \) on the set \( S = \{0, 1, 2, 3\} \). Concept: The operation \( a \times_4 b \) means multiplication modulo 4, i.e., \[ a \times_4 b = (a \times b) \mod 4 \] Solution: Step 1: Compute

Binary Operation on R x R 📺 Watch Video Explanation: Given: \( (a,b)*(c,d) = (a+c, b+d), \quad (a,b),(c,d)\in \mathbb{R}\times\mathbb{R} \) Commutativity: \( (a,b)*(c,d) = (a+c, b+d) = (c+a, d+b) = (c,d)*(a,b) \) ✔ Commutative Associativity: LHS: \( [(a,b)*(c,d)]*(e,f) = (a+c, b+d)*(e,f) = (a+c+e, b+d+f) \) RHS: \( (a,b)*[(c,d)*(e,f)] = (a,b)*(c+e, d+f) = (a+c+e, b+d+f) \)

Identity Element for HCF 📺 Watch Video Explanation: Find identity element Given: \( a * b = \mathrm{HCF}(a,b), \quad a,b \in \mathbb{N} \) Identity Definition: \( a * e = a \Rightarrow \mathrm{HCF}(a,e) = a \) Analysis: This implies: \( a \mid e \quad \text{for all } a \in \mathbb{N} \) So, \( e \)

Binary Operation on Ordered Pairs 📺 Watch Video Explanation: Given: \( (a,b)\circ(c,d) = (ac, bd), \quad (a,b),(c,d)\in \mathbb{R}_0 \times \mathbb{R}_0 \) i. Commutativity: \( (a,b)\circ(c,d) = (ac, bd) = (ca, db) = (c,d)\circ(a,b) \) ✔ Commutative Associativity: LHS: \( [(a,b)\circ(c,d)]\circ(e,f) = (ac,bd)\circ(e,f) = (ace, bdf) \) RHS: \( (a,b)\circ[(c,d)\circ(e,f)] = (a,b)\circ(ce,df) = (ace, bdf) \)